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Compute $\sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3}$ How to prove that
$$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad \frac{\pi^2G}{4}-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi^4}{64}+\frac{\Psi^{(3)}(\frac{1}{4})}{512}- \frac{\Psi^{(3)}(\frac{3}{4})}
{512}$$
This problem was proposed by @Ahmad Bow but unfortunately it was closed as off-topic and you can find it here.
Any way, I tried hard on this one but no success yet. here is what I did:
Using the identity
$$H_{n/2}=H_n-n\int_0^1 x^{n-1}\ln(1+x)\ dx, \quad x\mapsto x^2$$
$$H_{n/2}=H_n-2n\int_0^1 x^{2n-1}\ln(1+x^2)\ dx$$
We can write
$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\int_0^1\frac{\ln(1+x^2)}{x}\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}\ dx$$
where
\begin{align}
\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}&=\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^2}-\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^3}\\
&=\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^2}(1+(-1)^n-\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^3}(1+(-1)^n\\
&=\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^2}(1-(-1)^n-\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^3}(1-(-1)^n\\
&=\frac1{2x}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)
\end{align}
Therefore
$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\frac12\int_0^1\frac{\ln(1+x^2)}{x^2}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)\ dx$$
The sum can be done using the following identity
$$ \sum_{n=1}^{\infty} \frac{H_{n}}{ (n+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$
Differentiate both sides with respect to $a$ then set $a=1/2$ we get
$$\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}=\frac{45}{32}\zeta(4)-\frac74\ln2\zeta(3)$$
and the question here is how to calculate the the remaining integral or a different way to tackle the sum $S$ ? Thanks
| Using W.A.as well as my previous calculations on harmonics sums.
I find $$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad -\frac{\pi^2G}{4}-2G+\frac{7}{4}\zeta(3)-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi}{2}-\frac{\pi^2}{4}+\frac{\pi^3}{16}+\frac{\pi^4}{64}+\ln2-2-3\beta(4)+\frac{\Psi^{(3)}(\frac{1}{4})}{256}- \frac{\Psi^{(3)}(\frac{3}{4})}
{256}$$
$$ S=0,047743102114778065267...$$
But I'm not sure of the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Find all real values of the parameter a for which the equation $x^4+2ax^3+x^2+2ax+1=0$ has Find all real values of the parameter a for which the equation $x^4+2ax^3+x^2+2ax+1=0$ has
1) exactly two distinct negative roots
2) at least two distinct negative roots
I tried to factorize it but didn't get any breakthrough.
| Hint:
This is a reciprocal equation, so set $y=x+\dfrac1x$. Dividing the equation by $x^2$, the equation can be rewritten as
$$x^2+2ax+1+\frac{2a}x+\frac1{x^2}=x^2+\frac1{x^2}+2a\Bigl(x+\frac1x\Bigl)+1=y^2+2ay-1=0.$$
Now, as the reduced discriminant is $\Delta'=a^2+1>0$, this equation in $x$ has two roots with opposite signs, whichs are also the signs of $x$ (if $x$ is real):
$$y=-a\pm\sqrt{a^2+1\mathstrut}.$$
Can you continue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the inequality $\frac{n^3}{3} < 3n-3$ applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold I have been given the following task:
Prove that the following inequality applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold:
$$\frac{n^3}{3} < 3n-3$$
My idea was to prove the statement by using induction.
For $n=1$ it follows:
$\frac{1^3}{3} = \frac{1}{3} \nless 0 = 3 * 1-3$
For $n=2$ it follows:
$\frac{2^3}{3} = \frac{8}{3} < 3 = 3 * 2-3$
For $n=3$ it follows:
$\frac{3^3}{3} = \frac{27}{3} = 9 \nless 6 = 3 * 3-3$
I now assume that the inequality does not hold for $n \in \mathbb{N} \backslash \{2\}.$
n $\rightarrow$ n+1
$\frac{(n+1)^3}{3} = \frac{n^3+3n^2+3n+1}{3} = \frac{n^3}{3}+ \frac{3n^2+3n+1}{3} < 3n-3 + \frac{3n^2+3n+1}{3}=\frac{3(3n-3)+3n^2+3n+1}{3}=\frac{3n^2+12n-8}{3}$
But how do I continue from this step or is this even the wrong approach?
| Consider the opposite inequality:$$3n-3\leqslant\frac{n^3}3.$$As you proved, it holds for $n=3$. Now, assume that it holds for some $n\in\mathbb N\setminus\{1,2\}$. Then\begin{align}3(n+1)-3&=3n\\&=3n-3+3\\&\leqslant\frac{n^3}3+3\\&=\frac{n^3+9}3\\&\leqslant\frac{n^3+3n^2+3n+1}3\text{ (because $n>2$)}\\&=\frac{(n+1)^3}3.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3394427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$ The question is as the title says:
Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$.
I struggle to even start this question.
By inspection, I see that $x$ must be negative. Playing around yields $x=-3$ as a solution, though I do not know how to prove that there are no other solutions.
Upon differentiation, I obtain:
\begin{align}
\frac{d}{dx} (\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4})&= \frac{1}{3}(\sqrt[3]{(x+2)^{-2}}+\sqrt[3]{(x+3)^{-2}}+\sqrt[3]{({x+4})^{-2}})\\
\end{align}
which is always positive for all real values of $x$, implying that the function defined as $f(x)=\sqrt[3]{x+2} + \sqrt[3]{x+3} + \sqrt[3]{x+4}$ is strictly increasing.
Is there a better way to solve this equation?
| Let $f(x)=\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}$.
As $g(x)=x^3$ is increasing, the inverse function $h(x)=x^{1/3}$ is also increasing, so $f(x)$, the sum of 3 increasing functions, is also increasing. By the way, $f(x)$ is also continuous.
Inspection shows that $x=-3$ is a solution. As the function $f(x)$ is both continuous and increasing, $x=-3$ is the only real solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3394661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability of euclidean distance between two random points inside a unit circle/sphere greater than 1 Problem: Say there are two points inside the circle; A and B, and they are both randomly drawn according to a uniform distribution where the boundary is the circumference of the unit circle/the surface of the unit sphere. What's the probability that the euclidean distance between two randomly drawn points inside a unit circle/sphere greater than 1?
This question has two versions; the 2D one and the 3D one. I have almost gotten down the expression of the integral in 2D one, but I still get stuck at the late stage of the problem, I haven't tried the 3-D version just yet, but I guess I will get stuck at a similar stage.
The following is my attempt on the 2-D version of the problem:
Phase 1:
for the sake of simplicity, we can "fix" the angle θ of A to a particular fixed value θa and only vary its r value in the polar coordinate system, so A could be (0,θa), (0.2,θa), (1,θa) etc.
For B, we can vary everything including radius r and the angle θ of another point `B`.
Phase2:
The required probability should be equal to the sum of all of the conditional probability from r=0 to r= 1, where each increment of r is very very small:
ΣP{|A - B| > 1 | A= (r, θa) }
Upon taking this limiting process to the sum, this becomes a definite integral over the conditional probability density function from r=0 to r=1.
This is where I got stuck, I don't know how to transform the conditional probability to the conditional pdf inside the definite integral and possibly integrate it.
And after this 2-D version, the 3-D version is gonna be another beast that I need help in order to deal with that.
Note: These are the pictures of my drafts and my guesses, not sure whether they are helpful.
| This is the solution for the 2D problem
Fix $A$ to be a distance $r$ from the center of the circle (i.e. fix $|A| = r$). Note that the probability density function of $|A|$ is
$$f_{|A|}(r) = \frac{2\pi r}{\pi} = 2r \qquad r \in [0,1]$$
The $2\pi r$ is the "area" of the circle representing all possible $A$ for which $|A| = r$; and the $\pi$ in the denominator is the total area of the unit circle.
Given that $|A| = r$, what is the probability that $|A-B| <1$?
Draw a circle of radius $1$ centered at $A$, and find the area common to both circles. After a bit of geometry, you should find that the answer is $2\cos^{-1} \frac r2 - \frac{r\sqrt{1-r^2}}{2}$. Thus,
$$\Bbb P(|A-B|<1 \, \big| \, |A| = r) = \frac 1\pi \bigg(2\cos^{-1} \frac r2 - \frac{r\sqrt{1-r^2}}{2}\bigg)$$
Finally, we use the law of total probability:
\begin{align}
\Bbb P(|A-B|<1) & = \int_0^1 \Bbb P(|A-B|<1 \, \big| \, |A| = r) \cdot \color{red}{f_{|A|}(r)} \, dr \\
& = \int_0^1 \frac 1\pi \bigg(2\cos^{-1} \frac r2 - \frac{r\sqrt{1-r^2}}{2}\bigg) \cdot 2r \, dr \\
& = \frac 1\pi \int_0^1 \bigg(4r\cos^{-1} \frac r2 - r^2\sqrt{1-r^2}\bigg) \, dr \\
& = \frac 1\pi \bigg[ 2r^2 \cos^{-1} \frac r2 + 4\sin^{-1} \frac r2 - \frac 18 \sin^{-1} r - r\sqrt{4-r^2} + \frac 18 (r-2r^3)\sqrt{1-r^2} \bigg]^1_0 \\
& = \frac 1\pi \bigg(\frac{61}{48}\pi - \sqrt 3 \bigg) \\
& = \frac{61}{48} - \frac{\sqrt 3}{\pi}
\end{align}
And of course, what we really want is
$$\Bbb P(|A-B|>1) = 1 - \Bbb P(|A-B|<1) = \frac{\sqrt 3}{\pi} - \frac{13}{48} = 0.2805 \dots$$
As pointed out above in red, you are missing a factor of $2r$ in the integrand.
Here's how to do the integration
\begin{align}
& \int r\cos^{-1} \frac r2 \, dr \\
= & 4\int u\cos^{-1} u \, du && \text{substituted } u = \frac r2 \\
= & 4 \cdot \frac{u^2}{2} \cdot \cos^{-1} u - 4\int \frac{u^2}{2} \cdot \frac{-1}{\sqrt{1-u^2}} \, du && \text{integrated by parts} \\
= & 2u^2 \cos^{-1} u + 2\int u \cdot \frac{u}{\sqrt{1-u^2}} \, du \\
= & 2u^2 \cos^{-1} u + 2 \cdot u \cdot \big(-\sqrt{1-u^2}\big) - 2 \int 1 \cdot \big(-\sqrt{1-u^2}\big) \, du && \text{by parts again} \\
= & 2u^2 \cos^{-1} u - 2u\sqrt{1-u^2} + 2\int \sqrt{1-u^2} \, du \\
= & 2u^2 \cos^{-1} u - 2u\sqrt{1-u^2} + 2\int \sqrt{1-\sin^2 v} \cdot \cos v \, dv && \text{substituted } u = \sin v \\
= & 2u^2 \cos^{-1} u - 2u\sqrt{1-u^2} + 2\int \cos^2 v \, dv \\
= & 2u^2 \cos^{-1} u - 2u\sqrt{1-u^2} + 2 \bigg(\frac v2 - \frac{\sin (v) \cos (v)}{2} \bigg) \\
= & \frac 12 r^2\cos^{-1} \frac r2 - r\sqrt{1-\frac{r^2}{4}} + \sin^{-1} \frac r2 - \frac r2 \sqrt{1-\frac{r^2}{4}}
\end{align}
The other integral can be done by writing
$$\int r^2\sqrt{1-r^2} \, dr = \int r \cdot \Big(r\sqrt{1-r^2}\Big) \, dr$$
and integrating by parts as suggested.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3397331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Scalar potential in origo Find a scalar potential φ for the vector field F such that $\varphi(0,0)=5$ when
$$\mathbf{F}(x,y) = {3\cdot x\cdot \left(2\cdot y+1\right)}\mathbf{i}+{3\cdot x^2}\mathbf{j}$$
First we need to check that vector field is conservative so
$$\frac{\delta F_1}{\delta y}=\frac{\delta F_2}{\delta x}$$
$$6x=6x\to \frac{\delta F_1}{\delta y}-\frac{\delta F_2}{\delta x} = 0$$
then I want to find the a scalar potential $\varphi$ for the vector field $\mathbf F$ such that $\varphi(0,0)=5$.
So
$$f_x ={3\cdot x\cdot \left(2\cdot y+1\right)} \to f= 3\cdot x^2 \cdot y^2+\frac{3\cdot x^2}{2}+g(y) $$
then
$$f_y=3\cdot x^2+g'(y)=3\cdot x^2\to g(y)=k$$
and $$f(x,y)= 3\cdot x^2 \cdot y^2+\frac{3\cdot x^2}{2}+k$$
and $\varphi$ for the vector field $\mathbf F$ such that $\varphi(0,0)=5$.
$$3\cdot 0^2 \cdot 0^2+\frac{3\cdot 0^2}{2}+k=5 \to k=5$$ but the $k=5$ is wrong answer. I have done some stupid mistake but I can't see where.
| We have that
$$f_x ={3\cdot x\cdot \left(2\cdot y+1\right)} \implies f= 3\cdot x^2 \cdot y+\frac{3\cdot x^2}{2}+g(y)$$
$$f_y=3\cdot x^2+g'(y)=3\cdot x^2\implies g(y)=k$$
and therefore
$$f(x,y)= 3\cdot x^2 \cdot y+\frac{3\cdot x^2}{2}+k$$
but in any case $k=5$ for the given condition.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for all even positive integers $n = x^2 + 3y^2$ $(x, y \in \mathbb Z)$, $\dfrac{n}{4} = x'^2 + 3y'^2$ $(x, y \in \mathbb Z)$.
Prove that for all even positive integers $n$ that can be written in the form of $x^2 + 3y^2$ $(x, y \in \mathbb Z)$, $\dfrac{n}{4}$ can be also written in the form of $x'^2 + 3y'^2$ $(x', y' \in \mathbb Z)$.
It is not difficult to establish from the start that $4 \mid n \iff 2 \mid x + y$ and vice versa.
So $n$ either equals to $(2p)^2 + 3 \cdot (2q)^2$ or $(2p + 1)^2 + 3 \cdot (2q + 1)^2$ $(m, n \in \mathbb Z)$
$\implies \dfrac{n}{4}$ either equals to $p^2 + 3q^2$ or $p^2 + 3q^2 + p + 3q + 1$
For the first case, $\dfrac{n}{4} = x'^2 + 3y'^2$ if $x' = p$ and $y' = q$.
For the second case, $\dfrac{n}{4} = p^2 + 3q^2 + p + 3q + 1$. Since $4 \not\mid n \iff 2 \not\mid x + y$,
$\dfrac{n}{4}$ either equals to $$(2p')^2 + 3 \cdot (2q' + 1)^2 + (2p') + 3 \cdot (2q' + 1) + 1 = 4p'^2 + 12q'^2 + 2p' + 18q' + 7$$ $$ = (p' - 3q' - 2)^2 + 3(p' + q' + 1)^2$$ or $$(2p' + 1)^2 + 3 \cdot (2q')^2 + (2p' + 1) + 3 \cdot (2q') + 1 = 4p'^2 + 12q'^2 + 6p' + 6q' + 3$$ $$ = (p' - 3q')^2 + 3(p' + q' + 1)^2$$ $(p', q' \in \mathbb Z)$. I want to ask if the above solution is correct or not.
| The basic method you have used is correct but complications have been introduced by using $n$ for two different variables (see line 2 of your answer). A simple answer is as follows:-
If $n$ is even then, as in your proof, we can suppose $x$ and $y$ are both odd. Then $4$ is a factor of either $x+y$ or $x-y$ and, by changing the sign of $y$ if necessary, we can suppose $4$ is a factor of $x+y$.
Then simple algebra gives $\frac{n}{4}=\left(\frac{x-3y}{4}\right)^2+3\left(\frac{x+y}{4}\right)^2.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$?
It's known that $\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$.
Using the above statement, prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$.
My attempt
Obviously, we want to reach a statement such as $$\lim_{n \to \infty} \left(1 + \frac{-2}{n}\right)^n \quad \text{ or } \quad \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n \cdot \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n$$
in order to be able to apply the above condition. However, I was unable to achieve this. The furthest I've got was the following:
\begin{align}
\left(\frac{n-2}{3n+1}\right)^{2n} &= \left( \frac{9n^2 - 12n + 4}{9n^2 + 6n + 1} \right)^n\\
&= \left(1 + \frac{-18n+3}{9n^2+6n+1}\right)^n\\
f(n)&= \left(1 + \frac{-2 + \frac{3}{n}}{n+\frac{2}{3} + \frac{1}{9n}} \right)^n
\end{align}
It seems quite obvious that $\lim_{n \to \infty} \left(f(n)\right) = \left(1 + \frac{-2}{n + \frac{2}{3}}\right)^n$, however, this is not exactly equal to the statement given above. Are you able to ignore the constant and apply the condition regardness? If so, why? How would you go about solving this problem?
| Write
$$
\frac{3n-2}{3n+1} = 1-\frac{3}{3n+1}
$$
Recall that
$$
\left(1-\frac{3}{3n+1}\right)^{3n+1} \to e^{-3}
$$
Then
$$
\begin{align}
\left(1-\frac{3}{3n+1}\right)^{3n+1}
= &\left(1-\frac{3}{3n+1}\right)^{3n} \left(1-\frac{3}{3n+1}\right)
\\ \implies&
\left(1-\frac{3}{3n+1}\right)^{3n} \to e^{-3}
\\ \implies&
\left(1-\frac{3}{3n+1}\right)^{n} \to e^{-1}
\\ \implies&
\left(1-\frac{3}{3n+1}\right)^{2n} \to e^{-2}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405914",
"timestamp": "2023-03-29T00:00:00",
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Determine all $P(x) \in \mathbb R[x]$ such that $P(x^2) + x\big(mP(x) + nP(-x)\big) = \big(P(x)\big)^2 + (m - n)x^2$, $\forall x \in \mathbb R$.
Determine all polynomials $P(x) \in \mathbb R[x]$ knowing that $$ P(x^2) + x\big(mP(x) + nP(-x)\big) = \big(P(x)\big)^2 + (m - n)x^2, \forall x \in \mathbb R\,.$$
Replacing $x$ by $-x$, we have that $$P(x^2) - x[mP(-x) + nP(x)] = [P(-x)]^2 + (m - n)x^2, \forall x \in \mathbb R$$
Subtracting the second equation from the first, $$x(m + n)[P(x) + P(-x)] = P(x)^2 - P(-x)^2, \forall x \in \mathbb R$$
$$ \iff \left[ \begin{align} P(x) + P(-x) &= 0\\ P(x) - P(-x) &= x(m + n) \end{align} \right.$$
In the case of $P(x) +P(-x)=0$, we have that $$P(x^2) + (m - n)xP(x) = [P(x)]^2 + (m - n)x^2$$
In the case of $P(x) - P(-x) = x(m + n)$, we have that $$P(x^2) + x(m + n)[P(x) -nx] = [P(x)]^2 + (m - n)x^2$$
That neither of which can I answer is not unexpected.
| We conjecture that for odd degree $s\geq 3$ and above, the only polynomials that work are
$$P(x)=x^s+\frac{m-n}{2}x\text{ for }(m - n) (-2 + m - n)=0$$
while for even degree $s>3$ the only polynomials that work are
$$P(x)=x^s+\frac{m+n}{2}x\text{ for }(m - 3 n) (-2 + m + n)=0$$
In order to prove this, let
$$P(x)=\sum_{i=0}^s a_i x^i$$
where $a_s\neq 0$ and $s\geq 3$. Then by the equation given
$$0=P(x^2) + x (m P(x) + n P(-x))-P(x)^2-(m-n)x^2$$
$$=\sum_{i=0}^s a_i x^{2i}+\sum_{i=0}^s ma_i x^{i+1}+\sum_{i=0}^s na_i (-1)^ix^{i+1}-\sum_{i=0}^s\sum_{j=0}^sa_ia_jx^{i+j}-(m-n)x^2$$
We compute the highest coefficients first. Obviously, the highest power of $x$ is at most $2s$. This coefficient is
$$0=a_s-a_s^2\Rightarrow 0=a_s-1$$
This implies $a_s=1$. For $x^{2s-1}$, we have
$$0=-2a_{s-1}a_s=-2a_{s-1}$$
This implies $a_{s-1}=0$. For $x^{2s-2}$ we have
$$0=a_{s-1}-2a_{s-2}a_s-a_{s-1}^2=-2a_{s-2}$$
Again, this implies $a_{s-2}=0$. By induction, this pattern holds until $a_1$. It is this induction step that we use the fact that $s\geq 3$, else the pattern would break from $(m-n)x^2$. That is, we know
$$a_{s-1}=a_{s-2}=\cdots a_3=a_2=0$$
Then the polynomial is of the form
$$P(x)=x^s+ ax+b$$
Putting this into the equation gives us
$$0=(b-b^2)+x b (-2 a + m + n)+x^2(1-a) (a - m + n)+x^s(-2b)+x^{s+1}(m+n(-1)^s-2a)$$
Obviously, $b=0$. Then the equation becomes
$$0=x^2(1-a) (a - m + n)+x^{s+1}(m+n(-1)^s-2a)$$
Then
$$a=\frac{m+n(-1)^s}{2}$$
This gives us the first part of our conjecture. For the second part
$$0=(1-a) (a - m + n)=(2-m+n(-1)^s)(m+n(-1)^s-2m+2n)$$
Thus, either
$$2-m+n(-1)^s=0$$
or
$$m+n(-1)^s-2m+2n=0$$
For degree $0$, we have
$$P(x)=a$$
for some $a\in\mathbb{R}$. This implies
$$0=(a-a^2)+ax(m+n)+x^2(n-m)$$
Since this holds for all $x$, we know $a\in\{0,1\}$ and $m-n=0$. Without knowing more about $m$ and $n$, this is the best we can do. Now, let
$$P(x)=ax+b$$
for $a\neq 0$. Then
$$0=b(1-b)+xb (-2 a + m + n)+x^2(1 - a) (a - m + n)$$
This implies
$$b\in\{0,1\}$$
but other than than we can't say anything else without more restrictions on $m$ and $n$. Note that $(a,b)=(1,0)$ will work no matter what $n$ and $m$ are. For quadratics, let
$$P(x)=ax^2+bx+c$$
for $a\neq 0$. Then
$$0=c - c^2 + (-2 b c + c m + c n) x + (b - b^2 - 2 a c - m + b m + n -
b n) x^2 + (-2 a b + a m + a n) x^3 + (a - a^2) x^4$$
Since $a\neq 0$, this implies $a=1$. Then
$$0=c - c^2 + (-2 b c + c m + c n) x + (b - b^2 - 2 c - m + b m + n -
b n) x^2 + (-2 b + m + n) x^3$$
which implies
$$b=\frac{m+n}{2}\text{ and }c\in\{0,1\}$$
Then
$$P(x)=x^2+\frac{m+n}{2}x+c\text{ for }-2 c + \frac{1}{4} (m - 3 n) (-2 + m + n)=0\text{ with }c\in\{0,1\}$$
Having checked all degrees, let us discuss the results. First, if $m,n$ are arbitrary, then the only polynomial that works is $P(x)=x$. Second, we can list out all equations of $m$ and $n$ that give infinite solutions. We have
$$m - n=0$$
$$m - 3n=0$$
$$m+n=2$$
$$m-n=2$$
That is, if $m$ and $n$ do not satisfy one of these equations, then $P(x)$ has only finite solutions. The final equation which also gives a finite solution set is
$$-8 - 2 m + m^2 + 6 n - 2 m n - 3 n^2=0$$
which creates the quadratic
$$P(x)=x^2 + \frac{m + n}{2} x + 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
$\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$ $$\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$$
I tried the following but it doesn't seem to work...
$$= \lim_{x\to 0} \frac{x}{2} \cdot \frac{\sin(3x)}{3x}\cdot\frac{6x}{1-\cos(6x)}$$
$$= 0$$
But the result of this limit is $\frac{1}{6}$.
Am I missing something or did I make any glaring mistakes?
| More generally:
All you need is
$\dfrac{\sin(x)}{x} \to 1$
and
$1-\cos(x)
=2\sin^2(x/2)
$.
$\begin{array}\\
\dfrac{x\sin(ax)}{1-\cos(bx)}
&=ax^2\dfrac{\sin(ax)}{ax}\dfrac{1}{2\sin^2(bx/2)}\\
&=\dfrac{ax^2}{2(b/2)^2}\dfrac{\sin(ax)}{ax}\dfrac{(b/2)^2}{\sin^2(bx/2)}\\
&=\dfrac{a}{b^2/2}\dfrac{\sin(ax)}{ax}\dfrac{(bx/2)^2}{\sin^2(bx/2)}\\
&\to \dfrac{2a}{b^2}\\
\end{array}
$
For $a=3, b=6$
this is
$\dfrac{2\cdot 3}{6^2}
=\dfrac16
$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I solve the following inequality? I have the inequality:
$lg((x^3-x-1)^2) < 2 lg(x^3+x-1)$
And I'm not sure how should I go about solving it. I wrote it like this:
$2lg(x^3-x-1) < 2lg(x^3+x-1)$
$lg(x^3-x-1) < lg(x^3 + x - 1)$ (*)
Here I have the conditions:
$x^3-x-1 > 0$
$x^3+x-1 > 0$
At first this stumped me, but then I realised I can just add the inequalities to get:
$2x^3 -2> 0$
$x^3-1>0$
$x^3>1 \Rightarrow x \in (1, + \infty)$
So that is our condition.
Going back to (*) and raising the inequality to the power of $10$, I got:
$x^3-x-1 < x^3+ x - 1$
$2x>0 \Rightarrow x \in(0, +\infty)$
So, if we also consider the condition, we have
$x \in (1, + \infty) \cap (0, + \infty)$
So $x \in (1, + \infty)$.
The problem with this answer is that it is wrong. My textbook lists the following possible answers:
A. $\mathbb{R}$
B. $(0, + \infty)$
C. $(1, + \infty)$
D. $(0, 1)$
E. Other answer
So I got answer C, but after I checked the back of the book I found that the correct answer should be E. So, what did I do wrong and what is that other answer?
| Better not get rid of the squares -- they're your friend. So we have $$\log{(x^3-x-1)^2}< \log{(x^3+x-1)^2}.$$ Since the logarithm is monotonic, this implies $$(x^3-x-1)^2< (x^3+x-1)^2,$$ or that $$(x^3-x-1)^2- (x^3+x-1)^2<0.$$ This is easily factored to give $$(x^3-x-1+x^3+x-1)(x^3-x-1-x^3-x+1)<0,$$ or $$(2x^3-2)(-2x)<0,$$ or more simply $$x(x^3-1)=x(x-1)(x^2+x+1)>0,$$ which implies $$x(x-1)>0$$ since $x^2+x+1$ is always positive.
I believe you can now take it from here.
Well, you also have to factor in that the original inequality makes sense only for $x^3+x-1>0$ and $x^3-x-1\ne 0.$
| {
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In the range $0\leq x \lt 2\pi$ the equation has how many solutions $\sin^8 {x}+\cos^6 {x}=1$ In the range $0\leq x \lt 2\pi$ the equation has how many solutions
$$\sin^8 {x}+\cos^6 {x}=1$$
What i did
$\cos^6 {x}=1-\sin^8 {x}=(1-\sin^4 {x})(1+\sin^4 {x})=(1-\sin^2 {x})(1+\sin^2 {x})(1+\sin^4 {x})$
$\cos^4 {x}=(1+\sin^2 {x})(1+\sin^4 {x}) , \cos^2{x}=0$
$(1-\sin^2{x})^2=(1+\sin^2 {x})(1+\sin^4 {x})$
$-3\sin^2{x}=\sin^6{x}$
Which is not possilbe
Is there a trick or something to solve this equation or to know how many solutions are there ?
| We don't need any factorization, indeed since for $x\neq k\frac \pi 2$ we have that $0<|\sin x|<1$ and $0<|\cos x|<1$ then
$$\sin^8 {x}+\cos^6 {x}<\sin^2 {x}+\cos^2 {x}=1$$
we need only to check for the solutions $x= k\frac \pi 2$ which indeed are all the solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For triangle sides $a,b,c\in\Bbb N$, find semiperimeter $s$ if $\frac1{36}(s-a)^2+\frac19(s-b)^2+\frac14(s-c)^2=\frac1{49}s^2$ and $HCF(a,b,c)=1$
Let $a,b,c$ be the lengths of sides opposite to vertex $A$, $B$, and $C$ respectively in triangle $ABC$ . If
$$\frac{(s-a)^2}{36}+\frac{(s-b)^2}{9}+\frac{(s-c)^2}{4}=\frac{s^2}{49}$$
and $a,b,c \in \Bbb{N}$, and $\operatorname{HCF}(a, b, c)=1$. Then find the value of the semi-perimeter(s).
This is some a problem which involves some calculation. If we do some calculation with denominator, we get $36+9+4=49$, but I'm not getting the required result.
| We have the case of the equality in the inequality of Cauchy-Schwarz:
$$
s^2
=\left(
\frac{(s-a)^2}{6^2}+\frac{(s-b)^2}{3^2}+\frac{(s-c)^2}{2^2}
\right)
\left(6^2+3^2+2^2\right)
\ge
\Big(\ (s-a)+(s-b)+(s-c)\ \Big)^2
=s^2\ .
$$
So the two vectors $\Big(\ \frac 16(s-a),\ \frac 13(s-b),\ \frac 12(s-c)\ \Big)$ and $(6,3,2)$ are linearly dependent. We get for some scalar $k$, $s-a=36k$, $s-b=9k$, $s-c=4k$, so $s =(s-a)+(s-b)+(s-c)=49k$. We determine
$$
\begin{aligned}
a &=s-(s-a)=49k-36k=13k\ ,\\
b &=s-(s-b)=49k-9k=40k\ ,\\
c &=s-(s-c)=49k-4k=45k\ ,
\end{aligned}
$$
and we have to take $k=1$ now.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(1+a^2)(1+b^2)\geq 4ab$ for all $a,b>0$ without limits I'm facing a problem where I need to prove that $(1+a^2)(1+b^2)\geq 4ab$ for all $a,b>0$ without any use of limits.
I have tried 2 different methods that I'm not really sure about:
1) Taking out factors $a$ and $b$ so I get: $$ab(a+\frac{1}{a})(b+\frac{1}{b})\geq 4ab$$ $$(a+\frac{1}{a})(b+\frac{1}{b})\geq 4$$
and then split for cases: suppose $a,b<0.5$ so both $\frac{1}{a}>2$ and $\frac{1}{b}>2$
Therefore:$(a+\frac{1}{a})$>2 and $(b+\frac{1}{b})$>2 so $(a+\frac{1}{a})(b+\frac{1}{b}) >4$ and so on.
In cases both are smaller but very close to $1$ is there any general way to prove that $(a+\frac{1}{a})>2$?
2) The other method I've tried was more direct and algebric, but wasn't too helpfull:
$$(1+a^2)(1+b^2)=1+a^2+b^2+(ab)^2=1+(a-b)^2+2ab+(ab)^2$$
So I tried:
$$(1+a^2)(1+b^2)\geq4ab \rightarrow 1+(a-b)^2+(ab)^2\geq2ab$$
but without limits it's too hard to say anything about cases where $a\rightarrow0$ or $b\rightarrow0$ or both.
If there's anything I'm missing I'd like to be enlightened.
| Writing
$$1+a^2+b^2+a^2b^2\geq 4ab$$ and this is
$$a^2+b^2-2ab+1+a^2b^2-2ab\geq 0$$ and this is $$(a-b)^2+(ab-1)^2\geq 0$$
Or $$ab+\frac{1}{ab}+\frac{a}{b}+\frac{b}{a}\geq 2+2=4$$
since $$x+\frac{1}{x}\geq 2$$ if $$x\geq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$ If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the maximum value of $(x^2+y^2)^2$
My attempt is as follows:
$$(x-y)^2\ge 0$$
$$x^2+y^2\ge 2xy$$
$$2(x^2+y^2)\ge x^2+y^2+2xy$$
\begin{equation}
2(x^2+y^2)\ge (x+y)^2\tag{1}
\end{equation}
Solving the given equation:
$$x^2+2xy-y^2=6$$
$$(x+y)^2=2y^2+6$$
So putting the value of $(x+y)^2$ in equation $1$
$$2(x^2+y^2)\ge 2y^2+6$$
$$x^2+y^2\ge y^2+3$$
\begin{equation}
x^2\ge 3\tag{2}
\end{equation}
So $x\in \left(-\infty,-\sqrt{3}\right) \cup \left(\sqrt{3},\infty\right)$
But how to proceed from here?
| Use $$2(x^2+y^2)^2-(x^2+2xy-y^2)^2=(x^2-2xy-y^2)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how the following formula or any other explicit formula for computing Bernoulli numbers can be derived? how the following formula
$$B_n=1-\sum_{k=0}^{n-1}{n\choose k}\frac{B_k}{n-k+1}$$
can be derived?
I know how to use the formula but still have not seen any proof for the given formula.
|
The recurrence relation
\begin{align*}
B_n&=1-\sum_{k=0}^{n-1}{n\choose k}\frac{B_k}{n-k+1}\qquad n\geq 0\\
\end{align*}
specifies a variation of the Bernoulli numbers with $B_1=+\frac{1}{2}$ instead of the more common value $B_1=-\frac{1}{2}$.
We therefore start with the generating function
\begin{align*}
\sum_{k=0}^\infty \frac{B_k}{k!}x^k=\frac{x}{e^x-1}+x=\frac{xe^x}{e^x-1}\tag{1}
\end{align*}
Note: The recurrence relation also gives $B_0=1$ since we use the convention that an empty sum is equal to zero. This is the case for $n=0$ when the upper limit of the sum is less than the lower limit.
Multiplication of (1) by $e^x-1$ gives
\begin{align*}
xe^x&=\left(\sum_{k=0}^\infty\frac{B_k}{k!}x^k\right)\left(e^x-1\right)\tag{2}\\
&=\left(\sum_{k=0}^\infty\frac{B_k}{k!}x^k\right)\left(\sum_{l=1}^\infty\frac{x^l}{l!}\right)\tag{3}\\
&=\sum_{n=1}^\infty \left(\sum_{{k+l=n}\atop{k\geq 0,l\geq 1}}\frac{B_k}{k!}\frac{1}{l!}\right)x^n\tag{4}\\
&=\sum_{n=1}^\infty\left(\sum_{k=0}^{n-1}\frac{B_k}{k!}\frac{1}{(n-k)!}\right)x^n\tag{5}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{B_k}{k!}\frac{1}{(n-k+1)!}\right)x^{n+1}\tag{6}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}\frac{B_k}{n-k+1}\right)\frac{x^{n+1}}{n!}\tag{7}\\
\end{align*}
Comment:
*
*In (3) we expand the exponential function $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$.
*In (4) we do the Cauchy multiplication of series.
*In (5) we eliminate $l$ by using $l=n-k$.
*In (6) we shift the index $n$ to start with $n=0$.
*In (7) we expand numerator and denominator by $n!$ and introduce the binomial coefficient $\binom{n}{k}$.
From (7) and the expansion of the left-hand side of (2) we obtain after division by $x$
\begin{align*}
\sum_{n=0}^\infty \frac{x^n}{n!}=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}\frac{B_k}{n-k+1}\right)\frac{x^{n}}{n!}\tag{8}
\end{align*}
Comparison of the coefficient of $x^n$ gives after multiplication with $n!$ for $n\geq 0$:
\begin{align*}
\color{blue}{1}&=\sum_{k=0}^n\binom{n}{k}\frac{B_k}{n-k+1}\\
&\,\,\color{blue}{=B_n+\sum_{k=0}^{n-1}\binom{n}{k}\frac{B_k}{n-k+1}}
\end{align*}
and the claim follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving an inequality using AM-GM
Let $ a,b,c > 0$ such that $ abc = 1$. Prove that
$$\frac {ab}{a^2 + b^2 + \sqrt {c}} + \frac {bc}{b^2 + c^2 + \sqrt {a}} + \frac {ca}{c^2 + a^2 + \sqrt {b}}\le 1.$$
$$\frac{ab}{a^2+b^2+\sqrt{c}}=\frac{1}{\frac{a}{b}+\frac{b}{c}+\sqrt{c^3}}$$
(AM-GM): $$\frac{a}{b}+\frac{b}{a}+\sqrt{c^3}\geq 3\sqrt{c} $$
$$\frac{1}{\frac{a}{b}+\frac{b}{a}+\sqrt{c^3}}\leq \frac{1}{3\sqrt{c}}$$
Analogous: $$\frac{1}{\frac{b}{c}+\frac{c}{b}+\sqrt{a^3}} \leq \frac{1}{3\sqrt a}$$
$$\frac{1}{\frac{a}{c}+\frac{c}{a}+\sqrt{b^3}}\leq \frac{1}{3\sqrt b}$$
$$LHS \leq \frac{1}{3}+\frac{1}{3}+\frac{1}{3} =1$$
Am I right?
| Here is an easier way to prove the inequality. Just prove:
$$\frac {ab}{a^2 + b^2 + \sqrt {c}} \leq \frac {a+b}{2a + 2b + 2{c}} $$ and you are done! It is equivalent to $$a^2b+ab^2+2\leq a^3+b^3+ a\sqrt{c}+b\sqrt{c}$$ which is easy to see since $2\leq a\sqrt{c}+b\sqrt{c}$ and $ a^2b+ab^2\leq a^3+b^3$ are true by Am-Gm inequality
Edit to a comment. I tried to find such a number $k$ that a following is true:
$$\frac {ab}{a^2 + b^2 + \sqrt {c}} \leq \frac {a^k+b^k}{2a^k + 2b^k + 2c^k} $$
Playing with $a,b,c$ (I put $a=b=x$ and $c=1/x^2$) a find that this is possible only at $k=1$. This is a method I saw when trying to solve: https://artofproblemsolving.com/community/c6h17451p119168
| {
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Limits in multivariable Calculus $$\lim_{(x,y,z)\to(0,0,0)} \dfrac{(x+y+z)^2}{x^2 + y^2 + z^2}$$
According to me the answer should be $3$ but it says limit does not exist. Pls let me know where am I wrong.
The expression simplifies to $$1 + \dfrac{2\cdot(xy+yz+zx)}{x^2+y^2+z^2}$$
And for $x=y=z$ (say $x = y = z = a$) we have
$1+ 2\cdot \left(\dfrac{3a^2}{3a^2}\right)
= 1+2 =3$.
| You have determined the limit for the special case $x=y=z=t \to 0$ but for $x+y+z=0$, that is for example $x=t$, $y=-t-t^2$, $z=t^2$ we obtain
$$\frac{(x+y+z)^2}{x^2+y^2+z^2}=0$$
thereofore the limit doesn't exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\lim_{x \to 2} \frac{ x(x^2-1) }{x+3}=6/5$. Prove by $\epsilon$-$\delta$ definition that $\displaystyle\lim_{x \to 2} \frac{ x(x^2-1) }{x+3}=\dfrac{6}{5}$.
I know this is simple but I am stuck, If we assume $\delta <1$ and $|x-2|< \delta$ we can get rid of the factor $x-2$ in the following expression $$\dfrac{ x(x^2-1) }{x+3}-\dfrac{6}5=\dfrac{(x-2)(5x^2+10x+9)}{5x+15}$$
but I don't know how to proceed.
| From here
$$\left|\frac{ x(x^2-1) }{x+3}-6/5\right|=\left|\frac{(x-2)(5x^2+10x+9)}{5x+15}\right|$$
the trick is consider wlog $|x-2|<1 \iff 1<x<3$ and therefore
$$\left|\frac{(x-2)(5x^2+10x+9)}{5x+15}\right|=|x-2|\left|\frac{5x^2+10x+9}{5x+15}\right|\le \frac{84}{20}|x-2|\le5|x-2|$$
therefore it suffices to take $\delta =\frac{\epsilon}5$ to fulfill the definition, indeed in this case for any $x$ such that $0<|x-2|< \delta$
$$\left|\frac{ x(x^2-1) }{x+3}-6/5\right|\le 5|x-2|< 5 \cdot\frac \epsilon 5=\epsilon$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding maximum with two constraints Let $a$,$b$,$c$ be positive real numbers satisfying $$a+b+c=1$$ $$a^2+b^2+c^2=\frac{3}{8}$$ Find the maximum value of $$a^3+b^3+c^3$$
Using the well known $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, the problem changes to finding maximum of $abc$ but in this case AM-GM, cauchy schwarz or holder is not easy. Can someone help me? (I don't want lagrange multiplier solution)
| Let $a+b+c=3u,$ $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $u=\frac{1}{3}$, $$2(ab+ac+bc)=(a+b+c)^2-(a^2+b^2+c^2),$$ which gives
$$v^2=\frac{5}{48}.$$
Now, $$(a-b)^2(a-c)^2(b-c)^2\geq0$$ gives
$$3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6\geq0,$$ which gives
$$3uv^2-2u^3-2\sqrt{(u^2-v^2)^3}\leq w^3\leq3uv^2-2u^3+2\sqrt{(u^2-v^2)^3},$$
which says $$w^3\leq3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}=\frac{1}{32}.$$
Id est, $$a^3+b^3+c^3=27u^3-27uv^2+3w^3\leq27\cdot\frac{1}{27}-27\cdot\frac{1}{3}\cdot\frac{5}{48}+\frac{3}{32}=\frac{5}{32}.$$
The equality occurs for $a=b$, which gives $$(a,b,c)=\left(\frac{5}{12},\frac{5}{12},\frac{1}{6}\right),$$ which says that we got a maximal value.
| {
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Intersection of the blowing up and the exceptional curve for a concrete example Let $f(x,y)=x^2y+xy^2 - x^4-y^4=0$ be an affine curve.
Consider its blowing up at the origin, namely $x^2y+xy^2 = x^4+y^4$ and $xu=ty$ in $\mathbb{A}^2 \times \mathbb{P}^1$.
If $t \neq 0$ then we can set $y=xu$ and substituting this to the equation of curve, we obatin $x^3(2u-x(1+u^4))=0$, which implies the intersection of the blowing up and the exceptional curve is the point $u=0,x=0$. Because $f(x,y)=f(y,x)$, we get the same result for the case $ u \neq 0$ with $x=ty$, namely, we obtain $t=0,y=0$. Thus intersection of the blowing up and the exceptional curve is constructed by these two points. Is it correct?
Receive a comment.
I was wrong.
Case I.
$u=1$ and $x^2y+xy^2 = x^4+y^4$ and $xu=ty$ in $\mathbb{A}^2 \times \mathbb{P}^1$.
Then, $x^3\{u+u^2-x(1+u^4)\}=0$ and thus the intersection of exceptional curve $\{x=0\}$ and the blowing up is given by the equation $ u+u^2-0\times(1+u^4)=0$. Thus $u=0$ and $u=-1$ are the intersection.
Case II.
$t=1$ and $x^2y+xy^2 = x^4+y^4$ and $xu=ty$ in $\mathbb{A}^2 \times \mathbb{P}^1$.
Then, $y^3\{t+t^2-y(1+t^4)\}=0$ and thus the intersection of exceptional curve $\{y=0\}$ and the blowing up is given by the equation $ t+t^2-0\times(1+t^4)=0$. Thus $t=0$ and $t=-1$ are the intersection.
Because $(t:u)=(1:-1)=(-1:1)$, $t=-1$ in case II and $u=-1$ in case I are same points, thus the intersection of the blowing up and the exceptional curve consist three distinct points.
| As Lazzaro Campoetti says, you've made a mistake in your calculations. After substituting $y = xz$ and simplifying, I find the equation
$$
0 = x^3(z(1+z) - x(1+z^4)) \, .
$$
The factor $x^3$ corresponds to the exceptional locus. However, it is not a point, but a line: since $y = xz$, then $x = 0$ implies $y = 0$ and leaves $z$ unconstrained. Thus the exceptional locus is the $z$-axis, i.e., the vertical line $x = 0, y = 0$.
The other factor $z(1+z) - x(1+z^4)$ is the strict transform of the curve. Since $y = xz$, setting $z(1+z) - x(1+z^4) = 0$ we obtain the parametric equation
$$
(x,y,z) = \left(\frac{z(1+z)}{1+z^4}, \frac{z^2(1+z)}{1+z^4}, z \right) \, .
$$
To find the intersection with the exceptional locus, we must find the points where $x = 0, y = 0$. Since $y = xz$, then it suffices to solve $0 = x = \frac{z(1+z)}{1+z^4}$. Thus we find that the strict transform intersects the exceptional locus when $z = 0$ and $z = -1$. However, there is also one more intersection at infinity, since
$$
x = \frac{z(1+z)}{1+z^4} \to 0 \quad \text{as $z\to \infty$} \, .
$$
(You could also show this by setting $z = Z_1/Z_0$, clearing denominators, and then setting $Z_0 = 0$.) You can see these features in the plot in this SageMath cell.
Moreover, based on considerations of the tangent directions at the singular point $(0,0)$ of the original curve, you should expect there to be $3$ intersections. The coordinate $z = y/x$ keeps track of the tangent slopes, and as I mentioned in this comment, the tangent lines at the origin and $x = 0$, $y = 0$, and $y = -x$, which have slopes $\infty$, $0$, and $-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the value of $c$ to make $-\dfrac{1}{2}x^2+x+c$ a perfect square trinomial
I should find the value of $c$ to make: $$-\dfrac{1}{2}x^2+x+c$$ a
perfect square trinomial.
I really messed up: $-\dfrac{1}{2}x^2+x+c=x-\dfrac{1}{2}x^2+c$, but it seems like $-\dfrac{1}{2}x^2$ isn't $a^2\boldsymbol{-2ab}+b^2$. Can you help me?
| Extract the $-1/2$ from the equation:
$$ -\frac{1}{2}x^2 + x + c = -\frac{1}{2}( x^2 - 2x - 2c ) $$
The expression $ x^2 - 2x - 2c $ is similar to the expression $ x^2-2x + 1 = (x-1)^2 $ which gives that
$$ -2c = 1 \rightarrow c = -\frac{1}{2} $$
and the final expression is:
$$ -\frac{1}{2}(x-1)^2 $$
| {
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determining a prime factorization What is the prime factorizatíon of
$$N:= \sum_{n=1}^{10^8 -1} n^3 $$
writing the sum out, didn't help me..
therefore..any help is really appreciated!
| In general, $\sum_{n=1}^k n^3 =\frac {k^2(k+1)^2}4$
In your case $k=10^8 -1$
The sum is $\frac {(10^8-1)^2(10^8)^2}4$
$10^8-1=99999999=3^2 \times 11 \times 73 \times 101 \times 137$
$\frac {(10^8)^2}4=\frac {10^{16}}4=\frac{2^{16} \times 5^{16}}{2^2}=2^{14} \times 5^{16}$
So $\frac {(10^8-1)^2(10^8)^2}4=2^{14} \times 3^4 \times 5^{16} \times 11^2 \times 73^2 \times 101^2 \times 137^2 $
| {
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Simple continued fraction of $\sqrt{d}$ with period of shortest length $3$ This is the problem:
Does there exist positive integer $d$ ( which is not a perfect square ) such that the length of the least period in the simple continued fraction of $\sqrt{d}$ is $3$?
Consider the following theorem
Theorem : If the positive integer $d$ is not a perfect square, the simple continued fraction of $\sqrt{d}$ has the form $\sqrt{d} = [a_0;\overline{a_1,a_2,\cdots,a_{r-1},2a_o}]$ with $a_o = \lfloor d \rfloor$. Here $r$ denotes the length of the least period in the expansion of $\sqrt{d}$. Where $\lfloor x \rfloor$ denotes the greatest integer function/ floor function of $x$.
We want to solve for non-square $d$ where$\sqrt{d} = [a_0;\overline{a_1,a_2,2a_o}]$, and $a_o = \lfloor d \rfloor$. Since $d$ is a positive integer, $a_o = \lfloor d \rfloor \ge 1$, and $a_1 , a_2$ are positive integers by definition. Please note that the converse of the above theorem is not true, for example, consider $[1;\overline{1,1,2}] = \sqrt{10}/2$ and $[0;\overline{1,1,0}] = \sqrt{2}/2$. I calculated first few continued fractions for $\sqrt{d}$, $$\begin{array}{c|c|c} \sqrt{d} & \text{Continued fraction} & r\\ \hline
√2 & [1;\bar{2}] & 1 \\
√3 & [1;\overline{1,2}] & 2 \\
√5 & [2;\bar{4}] & 1\\
√6 & [2;\overline{2,4}] & 2\\
√7 & [2;\overline{1,1,1,4}] & 4\\
√8 & [2;\overline{1,4}] & 2\\
√10 & [3;\bar{6}] & 1\\
√11 & [3;\overline{3,6}] & 2\\
√12 & [3;\overline{2,6}] & 2\\
√13 & [3;\overline{1,1,1,1,6}] & 5\\
√14 & [3;\overline{1,2,1,6}] & 4\\
√15 & [3;\overline{1,6}] & 2\\
√17 & [4;\bar{8}] & 1\\
√18 & [4;\overline{4,8}] & 2\\
√19 & [4;\overline{2,1,3,1,2,8}] & 6 \\
√20 & [4;\overline{2,8}] & 2\\
√21 & [4;\overline{1,1,2,1,1,8}] & 6\\
√22 & [4;\overline{1,2,4,2,1,8}] & 6\\
√23 & [4;\overline{1,3,1,8}] & 4\\
√24 & [4;\overline{1,8}] & 2\\ \end{array}$$
As we can see for $1< d \le 24, r \ne 3$. Also, on a side note, observe that there does not exist two consecutive intergers $d$ and $d+1$ such that both $\sqrt{d}$ and $\sqrt{d+1}$ have $r=1$, moreover there are infinitely $\sqrt{d}$ such that the length of there least period is $1$ or $2$, $\sqrt{n^2+1} = [n;\overline{2n}]$, $\sqrt{n^2+2} = [n;\overline{n,2n}]$ and
$\sqrt{n^2-1} = [n-1;\overline{1,2(n-1)}]$ , where $n \in \mathbb{N}$ .Even for $r=4$, we have $\sqrt{n^2-2} = [n-1; \overline{1,n-2,1,2(n-1)}]$, $n>2$. Now i have a hunch that no such $d$ exists for which $\sqrt{d}$ have $r=3$. Any hints on how to prove this ? In general does there exist a number $m$ such that $r\ne m $ for any $\sqrt{d}$ ?
| Yes there are infinitely many. And it is not difficult to find them.
We seek continued fractions of the form
$\sqrt{N}=[a,\overline{b,c,2a}]$
First off, add $a$ to get a "pure" periodic expression. We shall call the quadratic surd $x$:
$x=a+\sqrt{N}=[\overline{2a,b,c}]$
We may then render
$x=2a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{x}}}$
and upon clearing fractions
$(bc+1)x^2+(b-c-2a(bc+1))x-(2ab+1)=0$
Now comes the sneaky part. If the above quadratic equation over the integers is to have a root $a+\sqrt{N}$, its other root must be $a-\sqrt{N}$ forcing the linear coefficient to be exactly $-2a$ times the quadratic one! Thereby $b=c$ above and the quadratic equation simplifies to:
$(b^2+1)x^2-2a(b^2+1)x-(2ab+1)=0$
This gives an integer radicand whenever $2ab+1$ is a multiple of $b^2+1$, in which cases the common factor of $b^2+1$ may be cancelled from the quadratic equation leaving the equation monic.
Suppose, for example, we drop in $b=2$. Then $2ab+1$ is to be a multiple of $5$ and $a$ can be any whole number one greater than a multiple of $5$. Putting $a=1$ results in the "trivial" solution $\sqrt{2}=[1,\overline{2}]$, as the period is reduced from three to one due to $b=c=2a$. But this equality is avoided for larger eligible values of $a$ and we get a series of period $3$ solutions. In all cases $N$ is one fourth the discriminant of the monic polynomial obtained after cancelling out the $b^2+1$ factor:
$a=6\to \sqrt{41}=[6,\overline{2,2,12}]$
$a=11\to \sqrt{130}=[11,\overline{2,2,22}]$
$a=5k+1\to \sqrt{25k^2+14k+2}=[5k+1,\overline{2,2,10k+2}]$
There are more families of solutions like this with other values of $b$. Just put in an even positive value for $b$ (why even?) and turn the crank. You must put $a>b/2$ to avoid the collapse we saw above with $\sqrt{2}$.
So much for a repeat petiod of $3$, what about larger periods?
Claim: For any positive whole numbers $r$ there are at least an infinitude of $\sqrt{N}$ continued fractions having repeat period $r$ where $N$ is a whole number, having the following form:
$\sqrt{N}=[kP_r+1;\overline{2,2,...,2,2(kP_r+1)}]$
$P_r$ is a Pell number defined by $P_0=0,P_1=P_{-1}=1,P_r=2P_{r-1}+P_{r-2}\text{ for } r\ge 2$, and $k$ is a whole number $\ge 0$ for $r=1$, $\ge 1$ otherwise. The number of $2$ digits before the final entries is $r-1$.
The proof bears some similarities to calculating the general solution for $r=3$ above. First add $kP_r+1$ to the expression to make a purely periodic fraction:
$x=kP_r+1+\sqrt{N}=[\overline{2(kP_r+1),2,2,...,2}]$
Then
$x=2(kP_r+1)+\dfrac{1}{[2,2,...,2,x]}$
By mathematical induction on $r$ and using the recursive relation defined for Pell numbers in the claim it is true that
$[2,2,...,2,x]=\dfrac{P_rx+P_{r-1}}{P_{r-1}x+P_{r-2}}$
with $r-1$ digits of $2$ in the block. When this is substituted into the previous equation the following is obtained:
$x=2(kP_r+1)+\dfrac{P_{r-1}x+P_{r-2}}{P_rx+P_{r-1}}$
$x=\dfrac{(2(kP_r+1)P_r+P_{r-1})x+2(kP_r+1)P_{r-1}+P_{r-2}}{P_rx+P_{r-1}}$
$(P_r)x^2-2(kP_r+1)P_rx-(2(kP_r+1)P_{r-1}+P_{r-2})=0$
Upon completing the square and back-substituting $\sqrt{N}=x-(P_rk+1)$ we obtain:
$N=\dfrac{(kP_r+1)^2P_r+2(kP_r+1)P_{r-1}+P_{r-2}}{P_r}$
Using the Pell number recursion to eliminate $P_{r-2}$:
$N=\dfrac{(kP_r+1)^2P_r+2(kP_r)P_{r-1}+P_r}{P_r}=(kP_r+1)^2+2kP_{r-1}+1$
thereby identifying $N$ as a whole number. For a full fundamental period $\ge 2$ the terminal element must not match the other elements, so in that case $k\ge 1$. Else (meaning a period of just $1$), $k$ may be any whole number, $k\ge 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving for terms of the sequence given by $ x^{2(2^k-1)}(x-1)=\ln2$ or gain some insight I'm interested in properties of the sequence given by (int k) :
$$ x^{2(2^k-1)}(x-1)= \ln2$$
$k = 1$: $x^2(x-1) = \ln2$ or $x^3-x^2 = \ln2$ :$x \approx 1.3737...$
$k=2$: $x^6(x - 1) = \ln2$ or $x^7 - x^6 = \ln2$ : $x\approx1.2175...$
$k=3$: $x^{14}(x-1) = \ln2$ or $x^{15} - x^{14} = \ln2$: $x\approx1.1284...$
Obviously this is asking for the base when certain consecutive powers differ by $\ln2$. Also apparent is how it approaches 1.
Unfortunately, I'm not able to solve them generally. Might there be some recursive property from one element to the next. Or maybe some series representation. Just looking for some insight and/or pointed in a good direction for what book I should crack open. Thanks in advance. ps. I'm just a novice hobbyist
| One can make approximations of the solution building, around $x=1$, the $[1,n]$ Padé approximant of
$$x^{2(2^k-1)}(x-1)- \log(2)$$
To fit in the page, I limited to $n=4$ and $a=\log(2)$. The approximation is
$$x=1+\frac{4 a^4 \left(2^k-2\right) \left(2^k-1\right) \left(2^{k+1}-3\right)+12 a^3
\left(-9\ 2^k+4^{k+1}+5\right)+36 a^2 \left(2^k-1\right)+6 a}{a
\left(2^k-1\right) \left(a \left(a \left(2^{k+1}-3\right) \left(a
\left(2^k-2\right) \left(2^{k+1}-5\right)+2^{k+5}-40\right)+18 \left(3\
2^{k+1}-7\right)\right)+48\right)+6}$$
This would give the following results
$$\left(
\begin{array}{ccc}
k & \text{approximation} & \text{exact} \\
1 & 1.36953 & 1.36955 \\
2 & 1.21408 & 1.21522 \\
3 & 1.12998 & 1.12815 \\
4 & 1.07960 & 1.07632 \\
5 & 1.04724 & 1.04507 \\
6 & 1.02664 & 1.02630 \\
7 & 1.01434 & 1.01516 \\
8 & 1.00747 & 1.00864 \\
9 & 1.00382 & 1.00486 \\
10 & 1.00193 & 1.00271
\end{array}
\right)$$
and I suppose that using this extimate as $x_0$, Newton method should converge in a couple of iterations.
| {
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What's the easiest way to find if this series is monotonic? This is my sequence:
$$a_n=\frac{3^n-4^n}{2^n-n+1}$$
So I am supposed to check how it is monotonic. Thats what i tried to run using wolfram, but honestly it gets complicated easily
$$\frac{3^{n+1}-4^{n+1}}{2^{n+1}-n}-\frac{3^n-4^n}{2^n-n+1}$$
I honestly don't know what to do after altering forms
| Let consider
$$a_n=\frac{3^n-4^n}{2^n-n+1}=\frac{3^n-4^n}{4^n}\cdot \frac{4^n}{2^n-n+1}=b_n \cdot c_n$$
and
$$b_n=\frac{3^n-4^n}{4^n}=\left(\frac34\right)^n-1$$
is negative strictly decreasing while
$$c_n=\frac{4^n}{2^n-n+1}$$
is positive strictly increasing indeed
$$\frac{4^{n+1}}{2^{n+1}-n}-\frac{4^n}{2^n-n+1}=4^n\left(\frac{4}{2^{n+1}-n}-\frac{1}{2^n-n+1}\right)>0$$
and
$$\frac{4}{2^{n+1}-n}-\frac{1}{2^n-n+1}>0\iff 4(2^n-n+1)> 2^{n+1}-n$$
$$2^{n+2}-2^{n+1}-3n+4>0 \iff 2^{n+1}-3n+4>0$$
which is true, therefore the product $b_n \cdot c_n$ is strictly decreasing.
| {
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Show that $\frac{a}{c} + \frac{b}{d} +\frac{c}{a} + \frac{d}{b}\le-12$ Let $a$,$b$,$c$ and $d$ be non-zero, pairwise different real numbers such that $ \frac{a}{b} +\frac{b}{c} +\frac{c}{d} + \frac{d}{a}=4$ and $ac=bd$ .
Show that $\frac{a}{c} + \frac{b}{d} +\frac{c}{a} + \frac{d}{b}\le-12$ and that $-12$ is the maximum.
I simplified the inequality to prove: $a^2+b^2+c^2+d^2\le -12ac$
But I am not sure what to do next.
Hints and solutions would be appreciated.
Taken from the 2018 Pan African Math Olympiad http://pamo-official.org/problemes/PAMO_2018_Problems_En.pdf
| The hint.
Prove that $$\frac{a}{b}+\frac{c}{d}=\frac{d}{c}+\frac{c}{d}\leq-2$$ or
$$\frac{b}{c}+\frac{d}{a}=\frac{a}{d}+\frac{d}{a}\leq-2$$ and use
$$\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=\left(\frac{a}{b}+\frac{c}{d}\right)\left(\frac{b}{c}+\frac{d}{a}\right).$$
A full solution.
Let $\frac{a}{b}+\frac{c}{d}>0$ and $\frac{b}{c}+\frac{d}{a}>0$.
Thus, by AM-GM $$4=\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right)=\left(\frac{d}{c}+\frac{c}{d}\right)+\left(\frac{a}{d}+\frac{d}{a}\right)\geq2+2=4,$$
which gives $c=d$ and $a=d$, which is impossible.
Thus, one of the expressions $\frac{a}{b}+\frac{c}{d}$ or $\frac{b}{c}+\frac{d}{a}$ is negative.
Let $u=\frac{a}{b}+\frac{c}{d}<0$.
Thus, $$u=\frac{a}{b}+\frac{c}{d}=\frac{d}{c}+\frac{c}{d}\leq-2$$ and we need to prove that
$$u(4-u)\leq-12$$ or
$$(u+2)(u-6)\geq0,$$ which is obvious.
The equality occurs for $\frac{a}{b}+\frac{c}{d}=-2$, $\frac{b}{c}+\frac{d}{a}=6$ and $ac=bd.$
Easy to see that it's possible, which says that $-12$ is a maximal value.
| {
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Sum of the given series: $(1^2 - 1 + 1)(1!) + (2^2 - 2 + 1)(2!) + \cdots + (n^2 - n + 1)(n!) $
Find the sum of the following series:
$$(1^2 - 1 + 1)(1!) + (2^2 - 2 + 1)(2!) + \cdots + (n^2 - n + 1)(n!) $$
I tried simplifying the $n^{th}$ term to use the method of telescoping to see if most of the terms get cancelled. But I couldn't simplify it in a way that was helpful. Any hints would be appreciated.
| We have
$$\sum_{k=1}^n (k^2-k+1)k!=\sum_{k=1}^n [(k+1)^2-3k]k!=\sum_{k=1}^n (k+1)(k+1)!-\sum_{k=1}^n kk!-2\sum_{k=1}^n kk!=$$
$$=(n+1)(n+1)!-1-2\sum_{k=1}^n kk!$$
and
$$\sum_{k=1}^n kk!=\sum_{k=1}^n (k+1-1)k!=\sum_{k=1}^n (k+1)!-\sum_{k=1}^n k!=(n+1)!-1$$
therefore
$$\sum_{k=1}^n (k^2-k+1)k!=(n+1)(n+1)!-1-2((n+1)!-1)=(n-1)(n+1)!+1$$
| {
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Why does this process map every fraction to the golden ratio? Start with any positive fraction $\frac{a}{b}$.
First add the denominator to the numerator: $$\frac{a}{b} \rightarrow \frac{a+b}{b}$$
Then add the (new) numerator to the denominator:
$$\frac{a+b}{b} \rightarrow \frac{a+b}{a+2b}$$
So $\frac{2}{5} \rightarrow \frac{7}{5} \rightarrow \frac{7}{12}$.
Repeating this process appears to map every fraction to
$\phi$ and $\frac{1}{\phi}$:
$$
\begin{array}{ccccccccccc}
\frac{2}{5} & \frac{7}{5} & \frac{7}{12} & \frac{19}{12} & \frac{19}{31} & \frac{50}{31} & \frac{50}{81} & \frac{131}{81} & \frac{131}{212} & \frac{343}{212} & \frac{343}{555} \\
0.4 & 1.40 & 0.583 & 1.58 & 0.613 & 1.61 & 0.617 & 1.62 & 0.618 & 1.62 & 0.618 \\
\end{array}
$$
Another example:
$$
\begin{array}{ccccccccccc}
\frac{11}{7} & \frac{18}{7} & \frac{18}{25} & \frac{43}{25} & \frac{43}{68} & \frac{111}{68} & \frac{111}{179} & \frac{290}{179} & \frac{290}{469} & \frac{759}{469} & \frac{759}{1228} \\
1.57143 & 2.57 & 0.720 & 1.72 & 0.632 & 1.63 & 0.620 & 1.62 & 0.618 & 1.62 & 0.618 \\
\end{array}
$$
Q. Why?
| Your numerators and denominators follow the same recursive relationship that defines the Fibonacci sequence. I.e. each time you make a new number (either a new numerator or a new denominator), the new number is equal to the sum of the two most recent previously made numbers.
Any sequence that follows this recursive relationship (the Fibonacci sequence being the most famous) has, as general term,
$$
x\cdot \varphi^n + y\cdot (1-\varphi)^n
$$
where the exact values of $x$ and $y$ are decided by what the first two numbers are.
Now note that $1-\varphi$ is a number between $-1$ and $0$, so $(1-\varphi)^n$ becomes really small as $n$ grows. Which is to say, your two numbers come closer and closer to being pure powerss of the golden ratio. And since they are (close to being) pure powers of the golden ratio, with exponents one apart, the ratio between them is (close to being) the golden ratio. This conclusion is valid for any starting point that doesn't give $x = 0$, which apart from starting at $\frac 00$ is impossible to do with integers.
| {
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Closed form of the Fibonacci sequence: solving using the characteristic root method Here is the official theorem I'll use:
Since the Fibonacci sequence is defined as $F_n=F_{n-1}+F_{n-2}$, we solve the equation $x^2-x-1=0$ to find that $r_1 = \frac{1+\sqrt 5}{2}$ and $r_2 = \frac{1-\sqrt 5}{2}$
So we have $F_n = c_1\left(\frac{1+\sqrt 5}{2}\right)^n + c_2\left(\frac{1-\sqrt 5}{2}\right)^n$
We know that $F_0 = F_1 = 1$. So we can solve the following system to find the values of $c_1$ and $c_2$:
$1 = c_1 + c_2$
$1 = c_1\left(\frac{1+\sqrt 5}{2}\right) + c_2\left(\frac{1-\sqrt 5}{2}\right)$
Solving this system does not give $c_1 = 1/\sqrt5, c_2 = -1/\sqrt 5$ , even though that is apparently the right answer, i.e. the closed form of the Fibonacci sequence is apparently $$\frac1{\sqrt 5}\left(\frac{1+\sqrt 5}{2}\right) -\frac1{\sqrt 5}\left(\frac{1-\sqrt 5}{2}\right)$$
Where did I go wrong? Why doesn't solving the system of equations give me $c_1 = 1/\sqrt5, c_2 = -1/\sqrt 5$?
| Let's see...
$$f_n = \left\{
\begin{array}{ll}
0 & \text{ for } n = 0 \\
1 & \text{ for } n = 1 \\
f_{n-1} + f_{n-2} & \text{ for } n>1
\end{array}
\right.$$
Now, the recursion can be written as $$f_n - f_{n-1} - f_{n-2} = 0,$$ so characteristic equation is $$x^2-x-1=0.$$
Now, the roots of the equation are $$X_{1,2} = \frac{1 \pm \sqrt{5}}2,$$
so general solution is
$$f_n = C_1\cdot\left(\frac{1 + \sqrt{5}}2\right)^n + C_2\left(\frac{1 - \sqrt{5}}2\right)^n$$
From the $f_1$ and $f_2$ we get
\begin{eqnarray}
0 &=& C_1 + C_2 \\
1 &=& C_1\left(\frac{1 + \sqrt{5}}2\right) + C_2\left(\frac{1 - \sqrt{5}}2\right)
\end{eqnarray}
From the first equation we get $$C_2 = -C_1,$$ so
\begin{equation}
1 = C_1\left(\frac{1 + \sqrt{5}}2\right) -C_1\left(\frac{1 - \sqrt{5}}2\right)
\end{equation}
Now, we have $$C_1\left[\frac{1 + \sqrt{5}}2 - \frac{1 - \sqrt{5}}2\right] = 1$$ or
$$C_1\cdot\sqrt{5} =1$$
So, $$C_1 = \frac{1}{\sqrt{5}}.$$
Now, $$C_2 = -\frac{1}{\sqrt{5}}.$$ The particular solution for the equation is therefore
$$f_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}2\right)^n - \left(\frac{1-\sqrt{5}}2\right)^n\right]$$
| {
"language": "en",
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Find $\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$. I have to find the limit:
$$\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$$
I tried multiplying with the conjugate of the formula:
$$(a-b)(a^2+ab+b^2)=a^3-b^3$$
So I got:
$$\lim\limits_{n \to \infty} \dfrac{n^3+2n^2+1-n^3+1}{\sqrt[3]{(n^3+2n^2+1)^2} + \sqrt[3]{(n^3+2n^2+1)(n^3-1)} + \sqrt[3]{(n^3-1)^2}}$$
$$\lim\limits_{n \to \infty} \dfrac{2n^2+2}{\sqrt[3]{n^6+4n^5+4n^4+2n^3+4n^2+1} + \sqrt[3]{n^6+2n^5-2n^2-1} + \sqrt[3]{n^6-2n^3+1}}$$
And I saw that we can factor $n^2$ in the denominator and if we do the same in the numerator, we'd get that the limit is equal to $2$. The problem is that my textbook claims that this limit is actually $\dfrac{2}{3}$. I don't see why should I have a $3$ in the denominator since the coefficient of $n^2$ would be $1$ if I would carry out the factoring to detail. So, what did I do wrong?
| The denominator is in the form
$$\sqrt[3]{n^6+...}+ \sqrt[3]{n^6+...}+ \sqrt[3]{n^6+...}\sim 3n^2$$
| {
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Prove derangement formula by induction I found this in my math book. I have solved a). Exercise b) is to prove the derangement sum by induction.
A derangement of $n$ elements is a permutation where none of the elements keep its original placement. Let $a_n$ be the number of possible derangements of n elements.
a) Show that $a_1=0$, $a_2=1$. Write all the derangements of the elements in $(A,B,C)$ and the elements in $(A,B,C,D)$. Show that the recursion formula is:
$a_n = (n-1)(a_{n-1} + a_{n-2})$
My answer: For placing element $1$ there are $(n-1)$ possibilities. If field $i$ does not take element 1, there is one forbidden element for each field, and there are $a_{n-1}$ possibilities left. If field $i$ does take element $1$, the problem is reduced to $a_{n-2}$. Because of that the formula is $a_n = (n-1)\left(a_{n-1} + a_{n-2}\right)$.
b) Show by induction that:
$a_n=n!\left[1 - \frac{1}{1!} + \frac{1}{2!} -... + (-1)^n\frac{1}{n!}\right]$.
My thoughts: I know how to prove it by the principle of inclusion and exclusion, but not induction. I guess the recursion formula from a) can be used.
| It is actually straightforward. It is clear that the formula holds for 1 and 2. Assume the formula holds for $1 \leq k \leq n $ and show that it holds for $n+1$
\begin{align}
a_{n+1} &= n \cdot (a_n + a_{n-1})\\
&= n \cdot ( n! \cdot [ 1 - \frac1{1!} + \cdots + (-1)^{n} \cdot \frac1{n!}] + (n-1)! \cdot [ 1 - \frac1{1!} + \cdots + (-1)^{n-1} \cdot \frac1{(n-1)!}] )\\
&= n \cdot ( (-1)^n + (n-1)! \cdot [ 1 - \frac1{1!} + \cdots + (-1)^{n-1} \cdot \frac1{(n-1)!}] \cdot (n+1)) \\
&= n \cdot (-1)^n + n \cdot (n+1) \cdot (n-1)! \cdot [ 1 - \frac1{1!} + \cdots + (-1)^{n-1} \cdot \frac1{(n-1)!}] \\
&= n \cdot (-1)^n + (-1)^n - (-1)^n + (n+1)! \cdot [ 1 - \frac1{1!} + \cdots + (-1)^{n-1} \cdot \frac1{(n-1)!}] \\
&= (-1)^n \cdot \frac{(n+1)!}{n!} - (-1)^n + (n+1)! \cdot [ 1 - \frac1{1!} + \cdots + (-1)^{n-1} \cdot \frac1{(n-1)!}] \\
&= (-1)^n \cdot \frac{(n+1)!}{n!} + (-1)^{n+1} \cdot \frac{(n+1)!}{(n+1)!}+ (n+1)! \cdot [ 1 - \frac1{1!} + \cdots + (-1)^{n-1} \cdot \frac1{(n-1)!}] \\
&= (n+1)! \cdot [ 1 - \frac1{1!} + \cdots + (-1)^{n+1} \cdot \frac1{(n+1)!}] \blacksquare \\
\end{align}
| {
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Find the $n^{th}$ derivative of $y=\dfrac {x^n}{x-1}$. Find the $n^{th}$ derivative of $y=\dfrac {x^n}{x-1}$.
My Attempt:
$$y=\dfrac {x^n}{x-1}$$
$$y=x^n\cdot(x-1)^{-1}$$
Differentiating both sides,
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdot n\cdot x^{(n-1)}$$
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdot n \cdot\dfrac {x^n}{x}$$
| $$\frac{x^n}{x-1}=\frac{x^n-1}{x-1}+\frac1{x-1}=P(x)+\frac1{x-1}$$ where $P$ is a polynomial of degree $n-1$. The $n^{th}$ derivative of this polynomial vanishes and we are left with the $n^{th}$ derivative of $\dfrac1{x-1}$,
$$\frac{(-1)(-2)\cdots(-n)}{(x-1)^{n+1}}.$$
| {
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General expression of polynomial roots sequence. Let $$P_n(X) = X^n-X^{n-1}-\cdots-1 = X^n - \sum_{k=0}^{n-1} X^k$$
The question is : Find the general expression of the sequence $(u_n)_{n \in \mathbb{N}}$ where $u_k$ is the greater root of $P_{k+1}(X)$.
We aldready know that $$u_0 = 1$$, $$u_1 = \frac{1+\sqrt{5}}{2}$$ and $$u_2 = \frac{1}{3}+\frac{1}{3} \sqrt[3]{19-3\sqrt{33}}+\frac{1}{3}\sqrt[3]{19+3\sqrt{33}}$$
We also can also think (by observations) that $u_h \to 2$ when $h \to \infty$.
But I can't find the general expression for $u_n$.
| By Descartes rule of signs, $P_n(x)$ has one sign change, so it has one root on the positive half of the real line. $P_n(2) = 1$ and $P_n \left( 2-\frac{1}{n} \right) = \frac{n - \left( 2-\frac{1}{n} \right)^n}{n-1}$. Since $n-1>0$ for $n > 1$, the sign of this latter expression is controlled by the numerator. If we can show the numerator is negative for $n$ sufficiently large, the root must lie in $(2-\frac{1}{n}, 2)$. As the lower end of this interval is approaching $2$, the limit of the only positive root of $P_n$ approaches $2$ as $n \rightarrow \infty$.
For $n > 2$, $2 - \frac{1}{n} > 3/2$, so $\left( 2 - \frac{1}{n} \right)^n > \left( \frac{3}{2} \right)^n$. We find that $\frac{\mathrm{d}}{\mathrm{d}n} \left( n - \left( \frac{3}{2} \right)^n \right) = 0 $ when $n = \frac{\ln(\ln 3 - \ln 2)}{\ln 2 - \ln 3}$ at which, $\frac{\ln(\ln 3 - \ln 2)}{\ln 2 - \ln 3} - \left( \frac{3}{2} \right)^{\frac{\ln(\ln 3 - \ln 2)}{\ln 2 - \ln 3}} = \frac{-(1+\ln\ln(3/2))}{\ln(3/2)} = -0.2399{\dots} < 0$. Therefore, the numerator above is negative (at least) for all $n > 2$.
| {
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"source": "stackexchange",
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Solve for $x,y$ $\begin{cases}\sin^2x+\sin^2y=\frac{3}{4}\\x+y=\frac{5\pi}{12}\end{cases}$
Designate $x,y\in\left(0,\frac{\pi}{2}\right)$ which fulfills
$\begin{cases}\sin^2x+\sin^2y=\frac{3}{4}\\x+y=\frac{5\pi}{12}\end{cases}$
My proof:
$\cos2x=1-2\sin^2x\Rightarrow \sin^2x=\frac{1}{2}-\frac{1}{2}\cos2x$
$\sin^2x+\sin^2y=\frac{3}{4}\Rightarrow1-\frac{1}{2}\left(\cos2x+\cos2y\right)=\frac{3}{4}\Rightarrow \cos2x+\cos2y=\frac{1}{2}\\ \cos x+\cos y=2\cos\frac{x+y}{2}\cos\frac{x-y}{2}\\\cos2x+\cos2y=\frac{1}{2}\Rightarrow \cos(x+y)\cos(x-y)=\frac{1}{4}\\\cos\left(\frac{5\pi}{12}\right)\cos\left(2x-\frac{5\pi}{12}\right)=\frac{1}{4}\\\cos2x=2\cos^2x-1\Rightarrow\cos\left(\frac{10\pi}{12}\right)=2\cos^2\left(\frac{5\pi}{12}\right)-1\Rightarrow\cos\left(\frac{5\pi}{12}\right)=\sqrt{\frac{-\frac{\sqrt{3}}{2}+1}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}\\\cos\left(2x-\frac{5\pi}{12}\right)=\frac{1}{4}\cdot\frac{2}{\sqrt{2-\sqrt{3}}}=\frac{1}{2\sqrt{2-\sqrt{3}}}$
I don't know what to do next,I don't know if it makes sense what I wrote
| From $\sin^2x+\sin^2y=\frac{3}{4}$, we get
$$\cos 2x +\cos 2y = \frac 12$$
or
$$\cos(x+y)\cos(x-y) = \frac14 \implies
\cos \frac {5\pi}{12}\cos(x-y) =\frac12\sin\frac\pi{6}$$
which leads to,
$$\cos(x-y) = \frac{\sin\frac\pi6}{2\cos\frac{5\pi}{12}}=\frac{2\sin\frac\pi{12}\cos\frac\pi{12}}{2\sin\frac\pi{12}}=\cos\frac\pi{12}$$
Thus,
$$x-y = \frac {\pi}{12}, \>\>\>\>\>x+y=\frac{5\pi}{12}$$
and
$$x= \frac\pi4,\>\>\>\>\>y= \frac\pi6$$
| {
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Prove:$\frac{1}{(x-1)^2+(y-1)^2}+\frac{1}{(x+1)^2+(y-1)^2}+\frac{1}{(x-1)^2+(y+1)^2}+\frac{1}{(x+1)^2+(y+1)^2}\geq2 $ ,if $-1
Prove:
$$\frac{1}{(x-1)^2+(y-1)^2}+\frac{1}{(x+1)^2+(y-1)^2}+\frac{1}{(x-1)^2+(y+1)^2}\\+\frac{1}{(x+1)^2+(y+1)^2}\geq2 $$
if $-1< x,y< 1$.
I tried Cauchy-Schwarz inequality and I proved Left side is greater or equal than 1
I also tried QM-GM inequality but I could not prove it.
| We need to prove that:
$$\left(\frac{1}{(x-1)^2+(y-1)^2}+\frac{1}{(x+1)^2+(y+1)^2}\right)+$$
$$+\left(\frac{1}{(x-1)^2+(y+1)^2}+\frac{1}{(x+1)^2+(y-1)^2}\right)\geq2$$ or
$$\left(\frac{1}{x^2+y^2+2-2(x+y)}+\frac{1}{x^2+y^2+2+2(x+y)}\right)+$$
$$+\left(\frac{1}{x^2+y^2+2-2(x-y)}+\frac{1}{x^2+y^2+2+2(x-y)}\right)\geq2$$ or
$$\frac{2(x^2+y^2+2)}{(x^2+y^2+2)^2-4(x+y)^2}+\frac{2(x^2+y^2+2)}{(x^2+y^2+2)^2-4(x-y)^2}\geq2$$ or
$$\frac{2(x^2+y^2+2)}{(x^2+y^2)^2+4-8xy}+\frac{2(x^2+y^2+2)}{(x^2+y^2)^2+4+8xy}\geq2$$ or
$$\frac{(x^2+y^2+2)(2(x^2+y^2)^2+8)}{((x^2+y^2)^2+4)^2-64x^2y^2}\geq1$$ or
$$(x^2+y^2)^3(2-x^2-y^2)+4(x^2+y^2)(2-x^2-y^2)+64x^2y^2\geq0,$$ which is obvious.
| {
"language": "en",
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"source": "stackexchange",
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Given the sequence $(a_n)_{n \ge 1}$ with $a_1=2$ and $a_{n+1} = \frac{n^2-1}{a_n} + 2$ for $n \ge 1$, find the following limits. I have the sequence $(a_n)_{n \ge 1}$, such that:
$$a_1 = 2, \hspace{1.5cm} a_{n+1} = \dfrac{n^2-1}{a_n}+2 \hspace{.25cm}, \forall n \ge 1$$
And I have to find $2$ limits:
$$\lim\limits_{n \to \infty} \dfrac{a_n}{n} \hspace{3.5cm} \lim\limits_{n \to \infty} \dfrac{\sum\limits_{k=1}^{n}a_k^3}{n^4}$$
The second one completly put me in the dark. I don't see any trick that I could use.
For the first one, I tried using Stolz-Cesaro:
$$\lim\limits_{n \to \infty} \dfrac{a_{n+1}-a_n}{n+1-n} = \lim\limits_{n \to \infty} (a_{n+1}-a_n) = \lim\limits_{n \to \infty} \bigg ( \dfrac{n^2-1}{a_n} - a_n \bigg )$$
And I got stuck. I don't think I can find a closed form for $a_n$, so I really don't know what should I do to find these $2$ limits.
| An easy proof by induction that $n\le a_n\le n+1$ with inductive step$$k\le a_k\le k+1\implies k+1=\frac{k^2-1}{k+1}+2\le a_{k+1}\le\frac{k^2-1}{k}+2<k+2$$implies the first limit is $1$ and the second is $\frac14$, since $\sum_{k=1}^nk^3=\frac14n^4+o(n^4)$. In particular, $1\le\frac{a_n}{n}\le1+\frac1n$ gets the first limit by the squeeze theorem, while$$\begin{align}\frac14\left(1+\frac1n\right)^2&=\frac{\sum_{k=1}^nk^3}{n^4}\le\frac{\sum_{k=1}^na_k^3}{n^4}\le\frac{\sum_{k=1}^n(k+1)^3}{n^4}\\&\le\frac{\sum_{k=1}^{n+1}k^3}{n^4}=\frac14\left(1+\frac1n\right)^2\left(1+\frac2n\right)^2\end{align}$$gives the second by the squeeze theorem. (Here I've used $\sum_{k=1}^nk^3=\frac13n^2(n+1)^2$.)
| {
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"source": "stackexchange",
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Factorize $xy(xy + 1) + (xy + 3) - 2(x + y + 1/2) - (x + y - 1)^2$ Factorize $xy(xy + 1) + (xy + 3) - 2(x + y + \frac{1}{2}) - (x + y - 1)^2$
When I substitute $x = 1$, $x = -1$, $y = 1$, $y = -1$, It gives a result of $0$. So the result will be $(x+1)(x-1)(y+1)(y-1)$ after checking.
However, it seems that the question pack these terms in a order for a reason(i.e. doesn't require student to expand). So what is the smart way to do it?
Any hints are welcomed. Thanks in advance.
| $$xy(xy+1)+(xy+3)-2(x+y+1/2)-(x+y-1)^2$$
$$xy(xy+1)+(xy+1)+2-2(x+y+1/2)-(x+y-1)^2$$
$$(xy+1)^2+2-2(x+y+1/2)-(x+y-1)^2$$
$$(xy+1)^2+2-2x-2y-1-(x+y-1)^2$$
$$(xy+1)^2-2(x+y-1)-1-(x+y-1)^2$$
$$(xy+1)^2-1-(x+y-1)(x+y+1)$$
$$(xy+1)^2-1-((x+y)^2-1)$$
$$(xy+1)^2-(x+y)^2$$
$$\Big[xy+1-(x+y)\Big]\Big[xy+1+(x+y)\Big]$$
$$\Big[y(x-1)+(1-x)\Big]\Big[y(x+1)+(1+x)\Big]$$
$$(x-1)(y-1)(x+1)(y+1)$$
| {
"language": "en",
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Closed expression for sum $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$ Inspired by the recent question if the series $\sum_{k=1}^{\infty} \frac{\sqrt{k}-\left\lfloor \sqrt{k}\right\rfloor}{k}$ diverges (which is the case) I became interested in the alternating series which is convergent by the Leibniz criterion.
The core of the problem is then the question if this sum
$$s = \sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}\simeq 0.591561$$
has a closed expression. Here $\left\lfloor {x}\right\rfloor$ is the greatest integer less than or equal to $x$.
I have found a nice integral representation for $s$ but I could not find a closed expression. Also, due to the slow convergence of the sum it is not trivial to get a numerical result with high accuracy which might be necessary to identify a possible closed expression.
Problems
a) find a closed expression for $s$
b) find the numerical result exact to 20 decimal places
| Update
We may use the convergence acceleration of alternating series developed by Cohen, Villegas, and Zagier.
Let
$$s = \ln 2 + \sum_{n=1}^\infty (-1)^n n
\sum_{i=1}^{n} \frac{1}{(n^2 + 2i-1)(n^2+2i)}$$
and
$$s_n = \ln 2 + \sum_{k=1}^n \frac{c_{n,k}}{d_n}\sum_{i=1}^k \frac{k}{(k^2 + 2i-1)(k^2+2i)}$$
where
\begin{align}
d_n &= \frac{(3+\sqrt{8})^n + (3-\sqrt{8})^n}{2}, \\
c_{n,k} &= (-1)^k \sum_{m=k+1}^n \frac{n}{n+m} \binom{n+m}{2m} 2^{2m}.
\end{align}
From Proposition 1 in [1], we have
$$|s-s_n| \le \frac{s}{d_n}.$$
Maple: $\mathrm{evalf}(s, 30) = 0.591560779349817340213846903345$,
$\mathrm{evalf}(s_{28} - s, 30) = 1.6944769437\cdot 10^{-21}.$
[1] Henri Cohen, Fernando Rodriguez Villegas, and Don Zagier,
"Convergence Acceleration of Alternating Series".
Previously written
We have
\begin{align}
s &= \sum_{k=1} (-1)^{k+1} \frac{\lfloor \sqrt{k}\rfloor}{k}\\
&= \sum_{n=1}^\infty \left(\sum_{k = n^2} (-1)^{k+1} \frac{\lfloor \sqrt{k}\rfloor}{k}\right)
+ \sum_{n=1}^\infty \left(\sum_{n^2 < k < (n+1)^2} (-1)^{k+1} \frac{\lfloor \sqrt{k}\rfloor}{k}\right)\\
&= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} - \sum_{n=1}^\infty (-1)^n n
\sum_{i=1}^{2n} (-1)^i \frac{1}{n^2 + i}\\
&= \ln 2 - \sum_{n=1}^\infty (-1)^n n
\sum_{i=1}^{2n} (-1)^{i} \frac{1}{n^2 + i}\\
&= \ln 2 + \sum_{n=1}^\infty (-1)^n n
\sum_{i=1}^{n} \frac{1}{(n^2 + 2i-1)(n^2+2i)}.\tag{1}
\end{align}
Maple can give numerical approximation of (1) with high accuracy. Or we may use "Convergence Acceleration of Alternating Series" technique to calculate (1).
| {
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"url": "https://math.stackexchange.com/questions/3450405",
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Show function is continuous at $(0,0)$ Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function with
$$\space f(x, y) = \begin{equation}
\begin{cases}
\dfrac{y^2\log(1+x^2y^2)}{\sqrt{x^4+y^4}}, & (x, y) \neq (0, 0)\\
0, & (x, y) = 0
\end{cases}\end{equation}$$
Show that f is continious.
I've already shown that f is continous for $(x, y) \neq (0,0)$ but I am having trouble with finding a proof for $(0, 0)$ using epsilon-delta or limits of sequences. Can anyone help?
This is how far I got with simplifying for epsilon delta
$$
\Bigg{|}\frac{y^2\log(1+x^2y^2)}{\sqrt{x^4+y^4}}\Bigg{|} \\
\Leftrightarrow \Bigg{|}\frac{y^2\log(1+x^2y^2)}{\sqrt{y^4*(\frac{x^4}{y^4}+1)}}\Bigg{|} \\
\Leftrightarrow \Bigg{|}\frac{y^2\log(1+x^2y^2)}{y^2\sqrt{(\frac{x^4}{y^4}+1)}}\Bigg{|} \\
\Leftrightarrow \Bigg{|}\frac{\log(1+x^2y^2)}{\sqrt{(\frac{x^4}{y^4}+1)}}\Bigg{|} \\
\leq \big{|}\log(1+x^2y^2)\big{|}
$$
But now I am clueless on how to connect
$$
\big{|}log(1+x^2y^2)\big{|} < \epsilon \\
$$
and
$$
\sqrt{x^2+y^2} < \delta
$$
| in the numerator $0\le y^2\log(1+x^2y^2)\le x^2y^4$
In the denominator $\sqrt{x^4+y^4} \ge \max (x^2,y^2)\ge 0$
| {
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Limit of this sequence $\lim_{n\to \infty}\frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}$ I am trying to calculate the limit of this sequence :
$$\lim_{n\to \infty}\frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}$$
I tried two methods and the two methods leaded me to infinity or 4/0.
Anything would be helpful , thanks.
| Call your limit $L$. Since $\sqrt{a+b}-\sqrt{a}=\frac{b}{\sqrt{a+b}+\sqrt{a}}$, $$L=\lim_{n\to\infty}\frac{\sqrt{2n+2}-\sqrt{2n-2}}{\sqrt{3n+1}-\sqrt{3n}}=\lim_{n\to\infty}\frac{4(\sqrt{3n+1}+\sqrt{3n})}{\sqrt{2n+2}+\sqrt{2n-2}}.$$Since $\sqrt{an+b}\sim\sqrt{an}$ for large $n>0$ and $a>0$,$$L=\lim_{n\to\infty}\frac{8\sqrt{3n}}{2\sqrt{2n}}=\frac{8\sqrt{3}}{2\sqrt{2}}=2\sqrt{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculation of the limit $\lim_{x\to 0} (\cos x)^{1/x^2}$ without De l'Hospital/Landau's symbols/asymptotic comparison I have calculate this limit
$$\lim_{x\to 0}\ (\cos x)^{1/x^2}$$
with these steps. I have considered that:
$$(\cos x)^{1/x^2}=(\cos x -1+1)^{1/x^2}=\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{x^2}}$$
I remember that $1/(\cos x -1)$ when $x\to 0$ the limit is $\infty$. Hence
$$\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{x^2}}=\left[\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{x^2}}\right]^{\frac{\frac{1}{\cos x -1}}{\frac{1}{\cos x -1}}}=\left[\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{\cos x-1}}\right]^{\frac{\frac{1}{x^2}}{\frac{1}{\cos x -1}}} \tag{1}$$
But if I take
$$p=\frac{1}{\cos x -1}\xrightarrow{x\to 0}p\to \infty$$
therefore I consider the
$$\lim_{p\to \infty}\left(1+\frac 1p\right)^p=e$$
Consequently for the $(1)$,
$$\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{\cos x-1}}\xrightarrow{p\to \infty} e$$
and the exponent
$$\lim_{x\to 0}\frac{\frac{1}{x^2}}{\frac{1}{-(-\cos x +1)}}=-\frac 12\tag{2}$$
At the end $\displaystyle \lim_{x\to 0}\ (\cos x)^{1/x^2}=e^{-\frac 12}$.
I have followed this strategy in my classroom with my students. Is there a shorter solution to the exercise than the one I have given?
| So you are assuming the fact $(1+u)^{1/u}\rightarrow e$ as $u\rightarrow 0$. With such, we also have $\dfrac{1}{u}\cdot\log(1+u)\rightarrow 1$, then
\begin{align*}
\dfrac{1}{x^{2}}\cdot\log(\cos x)&=\dfrac{1}{\cos x-1}\cdot\log(1+(\cos x-1))\cdot\dfrac{\cos x-1}{x^{2}}\\
&=\dfrac{1}{\cos x-1}\cdot\log(1+(\cos x-1))\cdot-2\cdot\dfrac{\sin^{2}\left(\dfrac{x}{2}\right)}{\left(\dfrac{x}{2}\right)^{2}}\cdot\dfrac{1}{4}\\
&\rightarrow-\dfrac{1}{2},
\end{align*}
so the limit goes to $e^{-1/2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Proving that $8^x+4^x\geq 5^x+6^x$ for $x\geq 0$. I want to prove that $$8^x+4^x\geq 6^x+5^x$$ for all $x\geq 0$. How can I do this?
My attempt:
I try by AM-GM: $$8^x+4^x\geq 2\sqrt{8^x4^x}=2(\sqrt{32})^x.$$
However, $\sqrt{32}\approx 5.5$ so I am not sure if $$2(\sqrt{32})^x\geq 5^x+6^x$$ is true.
Also, I try to compute derivatives but this doesn't simplify the problem. What can I do?
| The best and the easiest way to prove it is using Induction .
Base Case :
For $x=1$ , $ 8 + 4 \gt 6+5$
Induction Step :
Let us assume that $8^k +4^k \ge5^k + 6^k$ . This implies that $$8^k \ge (5^k + 6^k) - 4^k$$
Multiplying both the sides by $8$ , we get
$$\color{#f14}{8^{k+1} \ge 8\cdot5^k +8\cdot6^k -8\cdot4^k} \quad \quad \text{(1.)}$$
Now all that is left is to prove that the R.H.S is bigger than $5^{k+1}+6^{k+1}-4^{k+1}$.We prove it by assuming it is true . Then ,
$$\begin{align}8\cdot5^k +8\cdot6^k -8\cdot4^k & \ge 5^{k+1}+6^{k+1}-4^{k+1} \\
5^k(8-5) + 6^k(8-6) +4^k(4-8) & \ge 0 \\
3\cdot 5^k +2\cdot6^k - 4\cdot4^k & \ge 0 \\
\color{#2c0}{\left(\frac 34\right)\cdot \left(\frac 54\right)^k + \left(\frac 24\right)\cdot \left(\frac 64\right)^k }& \color{#2c0}{\ge 0 }\quad \quad \text{(2.)}\end{align}$$
which is obviously true for $k \ge 0$ . Hence our initial assumption was true.
Now combining $(1.)$ and $(2.)$ , we get ,
$$\color{#f14}{8^{k+1}} \ge \color{navy}{8\cdot5^k +8\cdot6^k -8\cdot4^k} \ge \color{#2c0}{5^{k+1}+6^{k+1}-4^{k+1}} $$
or $$ 8^{k+1} \ge 5^{k+1} + 6^{k+1} -4^{k+1} \implies \boxed { 8^{k+1} + 4^{k+1} \ge 5^{k+1} + 6^{k+1}}$$
Which completes our induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
Closed form of sum $\sum\limits_{k=1}^{\infty } \frac{(-1)^{k+1}}{\left\lfloor \sqrt{k}\right\rfloor}$ In two previous problems (Closed expression for sum $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$ and Closed expression for sum $\sum_{k = 1}^{\infty} \frac{\left\lfloor \sqrt{k} \right \rfloor}{k^2}$) the infinite sums contained the floor function (of a square root) in the numerator.
Here we ask, in a simple example, what happens if the floor function is in the denominator.
Question: what is the closed form of $\sum _{k=1}^{\infty } \frac{(-1)^{k+1}}{\left\lfloor \sqrt{k}\right\rfloor }$
| OK. Here's my generalization
which ran into some resistance
when posted as a separate problem.
Let $f(x)$ be such that
$f(1) = 1,
f'(x) > 0, f''(x) < 0,
f(x) \to \infty,
n \in \mathbb{N}
\implies f^{(-1)}(n)\in \mathbb{N}
$.
($f^{(-1)}(n)$
is the inverse function of $f$)
What can we say about
$$S=\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{\lfloor f(k) \rfloor} $$
Let $g$ be the
inverse function of $f$,
so
$f(g(x)) = g(f(x)) = x
$.
Let
$u(n) =
\begin{cases}
0 \text{ if } n \text{ odd}\\
1 \text{ if } n \text{ even}\\
\end{cases}
=\dfrac{(-1)^n+1}{2}.
$
\begin{align}
S
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{\lfloor f(k) \rfloor}\\
&=\sum_{n=1}^{\infty} \sum_{k=g(n)}^{g(n+1)-1} \dfrac{(-1)^{k+1}}{\lfloor f(k) \rfloor}\\
&=\sum_{n=1}^{\infty} \sum_{k=g(n)}^{g(n+1)-1} \dfrac{(-1)^{k+1}}{\lfloor n \rfloor}\\
&=\sum_{n=1}^{\infty} \dfrac1{n}\sum_{k=g(n)}^{g(n+1)-1} (-1)^{k+1}\\
&=\sum_{n=1}^{\infty} \dfrac1{n}\sum_{k=0}^{g(n+1)-g(n)-1} (-1)^{k+g(n)+1}\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{g(n)+1}}{n}\sum_{k=0}^{g(n+1)-g(n)-1} (-1)^{k}\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{g(n)+1}}{n}u(g(n+1)-g(n)-1)\\
\end{align}
If $f(k) = \sqrt{k}$,
then
$g(n) = n^2$
so
$u(g(n+1)-g(n)-1)
=u(2n)
=1
$
and
$(-1)^{g(n)+1}
=(-1)^{n^2+1}
=(-1)^{n+1}
$
so
$$S
=\sum_{n=1}^{\infty} \dfrac{(-1)^{g(n)+1}}{n}u(g(n+1)-g(n)-1)
=\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}
=\ln(2).
$$
If $f(k) = \sqrt[3]{k}$,
then
$g(n) = n^3$
so
$u(g(n+1)-g(n)-1)
=u(3n^2+3n)
=u(3n(n+1))
=1
$
and
$(-1)^{g(n)+1}
=(-1)^{n^3+1}
=(-1)^{n+1}
$
so
$$S
=\sum_{n=1}^{\infty} \dfrac{(-1)^{g(n)+1}}{n}u(g(n+1)-g(n)-1)
=\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}
=\ln(2).
$$
If $f(k) = \sqrt[m]{k}$,
then
$g(n) = n^m$
so
$\begin{array}\\
u(g(n+1)-g(n)-1)
&=u((n+1)^m-n^m-1)\\
&=u(\sum_{j=1}^{m-1} \binom{m}{j}n^j)\\
&=u(\sum_{j=1}^{\lfloor \frac{m-1}{2} \rfloor} (\binom{m}{j}n^j+\binom{m}{m-j}n^{m-j})
\qquad\text{central binomial coefficient is even}\\
&=u(\sum_{j=1}^{\lfloor \frac{m-1}{2} \rfloor} (\binom{m}{j}(n^j+n^{m-j}))\\
&=u(\sum_{j=1}^{\lfloor \frac{m-1}{2} \rfloor} (\binom{m}{j}n^j(1+n^{m-2j}))\\
&=1
\qquad\text{since }n^j(1+n^{m-2j})
\text{ is even}\\
\end{array}
$
and
$(-1)^{g(n)+1}
=(-1)^{n^m+1}
=(-1)^{n+1}
$
so
$$S
=\sum_{n=1}^{\infty} \dfrac{(-1)^{g(n)+1}}{n}u(g(n+1)-g(n)-1)
=\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}
=\ln(2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solve without L'Hopital's rule: $\lim_{x\to0}\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}$ Solve without L'Hopital's rule:
$$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}$$
My work:
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}=\displaystyle\lim_{x\to0}{\frac{\cos{(2x)}\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\sin{(2x)}}}$
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\sin{2x}}}=\displaystyle\lim_{x\to0}{\frac{\cosh(3x^2)\cdot e^{8x^3}-1}{x^2\sin{(2x)}}}\cdot\displaystyle\lim_{x\to0}{\frac{1}{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}+1}}$
All of my attempts failed.
$$\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}+1\;\;\text{is continuous &decreasing }$$
Source: Matematička analiza 1, 2. kolokvij
| \begin{align*}
\dfrac{\sqrt{\cosh(3x^{2})}e^{4x^{3}}-1}{x^{2}\tan(2x)}&=\dfrac{\cosh(3x^{2})e^{8x^{3}}-1}{x^{2}(2x)}\dfrac{2x}{\tan(2x)}\dfrac{1}{\sqrt{\cosh(3x^{2})}e^{4x^{3}}+1},
\end{align*}
so the issue is to compute
\begin{align*}
\lim_{x\rightarrow 0}\dfrac{\cosh(3x^{2})e^{8x^{3}}-1}{x^{2}(2x)}.
\end{align*}
Note that
\begin{align*}
\dfrac{\cosh(3x^{2})e^{8x^{3}}-1}{x^{2}(2x)}=\dfrac{(e^{3x^{2}}+e^{-3x^{2}})e^{8x^{3}}-2}{2(2x^{3})}.
\end{align*}
If you are allowed to use power series, then
\begin{align*}
(e^{3x^{2}}+e^{-3x^{2}})e^{8x^{3}}-2\approx 16x^{3},
\end{align*}
so the cancellation with the $x^{3}$ in the denominator is safe.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to find matrix $B$?
let $A$ be a $2 \times 2$ matrix and $I$ be the identity matrix . Assume that the null spaces of $A-4I$ and $A-I$ respectively spanned by $\begin{bmatrix} 3 \\ 2\end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1\end{bmatrix}$ respectively.Find a matrix $B $ such that $B^2= A$
My attempt :
According to question , it tell that $Au= 4u$ and $Av= v$
Now take u = $\begin{bmatrix} 3 \\ 2\end{bmatrix}$ and $v = \begin{bmatrix} 1 \\ 1\end{bmatrix}$
$A\begin{bmatrix} 3 \\ 2\end{bmatrix}=4\begin{bmatrix} 3 \\ 2\end{bmatrix}$
$A\begin{bmatrix} 1 \\ 1\end{bmatrix}= \begin{bmatrix} 1 \\ 1\end{bmatrix}$
After that im not able to proceed further
any hints/solution
| You know that $A.\left[\begin{smallmatrix}3\\2\end{smallmatrix}\right]=4\left[\begin{smallmatrix}3\\2\end{smallmatrix}\right]$ and that $A.\left[\begin{smallmatrix}1\\1\end{smallmatrix}\right]=\left[\begin{smallmatrix}1\\1\end{smallmatrix}\right]$. So, if $P=\left[\begin{smallmatrix}3&1\\2&1\end{smallmatrix}\right]$, then$$P^{-1}.A.P=\begin{bmatrix}4&0\\0&1\end{bmatrix}=\begin{bmatrix}2&0\\0&1\end{bmatrix}^2.$$So, take $B=P.\left[\begin{smallmatrix}2&0\\0&1\end{smallmatrix}\right].P^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Function that requires differentiation Differentiate:
$$\ln\left(\dfrac{x^2\sqrt{2x^2+3}}{\left(x^4+x^2\right)^3}\right)$$
I have tried to figure it out here:
The steps are too long so I tidy up as an image
After the steps from the image, these are the final steps of simplifying:
$$=\dfrac{\frac{2x^5\left(x^2+1\right)^3}{\sqrt{2x^2+3}}-\sqrt{2x^2+3}\left(4x^3\left(x^2+1\right)^3+6x^5\left(x^2+1\right)^2\right)}{x^4\left(x^2+1\right)^3\sqrt{2x^2+3}}$$
$$=\dfrac{x^4\left(x^2+1\right)^3\left(-\frac{4\sqrt{2x^2+3}}{x^5\left(x^2+1\right)^3}-\frac{6\sqrt{2x^2+3}}{x^3\left(x^2+1\right)^4}+\frac{2}{x^3\left(x^2+1\right)^3\sqrt{2x^2+3}}\right)}{\sqrt{2x^2+3}}$$
$$=\dfrac{2x}{2x^2+3}-\dfrac{6x}{x^2+1}-\dfrac{4}{x}$$
Please tell me whether this is correct or not, I would like to simplify my steps further if I could. Thanks!
| To simplify we can use that
$$f(x)=\ln\left(\dfrac{x^2\sqrt{2x^2+3}}{\left(x^4+x^2\right)^3}\right)=\ln x^2+\ln\sqrt{2x^2+3}-\ln \left(x^4+x^2\right)^3=$$
$$=2\ln x+\frac12\ln(2x^2+3)-3\ln(x^4+x^2) $$
and therefore
$$f'(x)=2\cdot \frac1x+\frac12\cdot\frac{4x}{2x^2+3}-3\cdot\frac{4x^3+2x}{x^4+x^2}=\frac 2x+\frac{2x}{2x^2+3}-\frac{12x^2+6}{x(x^2+1)}$$
which is equivalent to the expression you have found indeed
$$\frac 2x+\frac{2x}{2x^2+3}-\frac{12x^2+6}{x(x^2+1)}=\frac{2x}{2x^2+3}-\frac{12x^2+6-2(x^2+1)}{x(x^2+1)}=$$
$$=\frac{2x}{2x^2+3}-\frac{10x^2+4}{x(x^2+1)}=\frac{2x}{2x^2+3}-\frac{4(x^2+1)}{x(x^2+1)}-\frac{6x^2}{x(x^2+1)}=$$
$$=\frac{2x}{2x^2+3}-\frac{4}{x}-\frac{6x}{x^2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the limit of $\sqrt[n]{n^2 + n}$ To find the limit I got the $\sqrt[3n]{n^2+n}$
Particularly, $\sqrt[3n]{n^2+n} \ge 1 \rightarrow \sqrt[3n]{n^2+n} = 1 + d_n$ where $d_n\ge 0$.
According to the Bernoulli's rule
$\sqrt{n^2+n} = (1+d_n)^n \ge d_n\cdot n \rightarrow d_n \le \frac{\sqrt{n^2+n}}{n}$
The $\frac{\sqrt{n^2+n}}{n} \rightarrow 1$, so $\lim d_n=1 $
So, $\lim\sqrt[n]{n^2+n} = \lim (1+d_n)^3 = \lim(1+3d_n^2+3d_n+d_n^3) =8$
However, $\sqrt[n]{n^2+n}$ tends to $1$. Where is the problem of my solution ? Can you give me a hint of how can I solve it with Bernoulli's rule ?
| Option:
$1\le (n^2+n)^{1/n} \le (2n^2)^{1/n} =$
$2^{1/n}(n)^{1/n}(n)^{1/n}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
number of such real values of $a.$
Let $a$ be a real number in the interval $[0,314]$ such that
$$\displaystyle \int^{3\pi+a}_{\pi+a}|x-a-\pi|\sin \frac{x}{2}dx=-16.$$
Determine the number of such real values of $a.$
What I tried:
Put $x-a-\pi=t$. Then,
$$\displaystyle \int^{2\pi}_{0}|t|\sin \bigg(\frac{t+a+\pi}{2}\bigg)dx=-16$$
$$\int^{2\pi}_{0}t\cos\bigg(\frac{t+a}{2}\bigg)dt=-16$$
$$2t\sin \frac{t+a}{2}\bigg|^{2\pi}_{0}+4\int^{2\pi}_{0}\cos\frac{a+t}{2}dt=-16$$
$$-2\pi\sin \frac{a}{2}+4\int^{2\pi}_{0}\cos\frac{a+t}{2}dt=-16.$$
How do I solve this?
| You already have
$$\int^{2\pi}_{0}t\cos\bigg(\frac{t+a}{2}\bigg)dt=-16$$
From here, we have
$$2t\sin\frac{t+a}{2}\bigg|^{2\pi}_{0}-2\int_0^{2\pi}\sin\frac{t+a}{2}dt=-16$$
$$\iff -4\pi\sin\frac a2-2\int_0^{2\pi}\bigg(\sin\frac t2\cos\frac a2+\cos\frac t2\sin\frac a2\bigg)dt=-16$$
$$\iff -4\pi\sin\frac a2-2\cos\frac a2\int_0^{2\pi}\sin\frac t2dt-2\sin\frac a2\int_0^{2\pi}\cos\frac t2dt=-16$$
$$\iff -4\pi\sin\frac a2-8\cos\frac a2=-16$$
$$\iff \pi\sin\frac a2+2\cos\frac a2=4\tag1$$
The LHS of $(1)$ can be written as
$$\sqrt{\pi^2+2^2}\sin\bigg(\frac a2+b\bigg)$$
for some $b\in\mathbb R$.
Now, we have
$$\sqrt{\pi^2+2^2}\sin\bigg(\frac a2+b\bigg)\le \sqrt{\pi^2+4}\lt\sqrt{(2\sqrt 3)^2+4}=4$$
because of $\pi\lt 2\sqrt 3$. (See here) So, the LHS of $(1)$ is smaller than $4$.
It follows that the number of $a$ is $\color{red}0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof of $f(x) = \sum_{k=0}^\infty a_kx^k$ with $a_0 := a_1 := 1$ and $a_{n+2} := a_n+a_{n+1}$ A recursion is defined by $a_0 := a_1 := 1$ and $a_{n+2} := a_n+a_{n+1}$ for $n \geq 2$.
How can one prove that the power series
$$f(x) = \sum_{k=0}^\infty a_kx^k$$ converges absolutely for $|x| < \frac{1}{2}$ and
$$f(x) = - \frac{1}{x^2+x-1} \text{ for } |x| < \frac{1}{2}$$
How can one conclude from what is given above that
$$a_{n-1} = \frac{1}{\sqrt{5}} \big( \big( \frac{1+\sqrt{5}}{2} \big)^n - \big ( \frac{1-\sqrt{5}}{2} \big )^n\big) \forall n \in \mathbb{N}$$
I have found this answer on another question. The problem is that it doesn't answer the $|x| < \frac{1}{2}$ part and the conclusion.
The reciprocals of the poles of the denominator of this rational
function are the roots of $X^2-X-1$, which are the golden ratio $\phi$
and its "conjugate" $-\phi^{-1}$. The largest in absolute value of
these two is $\phi$, so the coefficients $c_n$ of your series are
asymptotically $\sim c\phi^n$ for some $c\neq0$, and the radius of
convergence is $\frac1\phi$.
You make an interesting observation that this convergence condition is
not equivalent to $|x+x^2|<1$. Let's see what happens when you expand
$\sum_{i=0}^\infty (x+x^2)^i$. You get
$1+x+2x^2+3x^3+5x^4-8x^5+\cdots$, which looks familiar; indeed it is
$\sum_iF_{i+1}x^i$ where $F_i$ are the Fibonacci numbers (which indeed
grow asymptotically as $\frac1{\sqrt5}\phi^n$). Now what happens if
one sets $x=-1$, a value outside the radius of convergence, but which
conveniently satisfies $|x+x^2|=0<1$? On one hand you have been
expanding $\sum_{i=0}^\infty 0^i$ which should give $1$. On the other
hand substituting $x=-1$ into the series gives $1-1+2-3+5-8+\cdots$
which diverges. What gives? Well if you try to find your powers of
$x+x^2$ back in the power series, you can group together parts of
several terms so that each group of terms (except the initial $1$)
evaluates to $0$ for $x=-1$. But in order to do so, you have been, at
$x=-1$, grouping together terms in an absolutely divergent series, and
that is not allowed (and may produce a convergent series). Indeed an
easier form of this swindle is $$ \begin{align} 1-1+2-3+5-8-\cdots
> &= 1 - (0+1)+(1+1)-(1+2)+(2+3)-(3+5) \\ &=
> 1+(-1+1)+(1-1)+(-2+2)+(3-3)+(-5+\cdots \\ &=1. \end{align} $$
Your power series is divergent at$~x=-1$ (and indeed everywhere
outside the interval $[-\frac1\phi,\frac1\phi]$).
| Note that $|a_n| \le 2^k$, hence ${1 \over R} = \limsup_n \sqrt[n]{|a_n|} \le 2$.
To obtain the functional form we have $a_{n+2} = a_{n+1} + a_n$, so
$\sum_{n=0}^\infty a_{n+2} x^{n+2} = \sum_{n=0}^\infty a_{n+2} x^{n+2} + \sum_{n=0}^\infty a_{n+2} x^{n+2} $ from which we get
$f(x)-1-x = x(f(x)-1) + f(x) x^2$, and from this we get the desired form of $f$.
Let $\lambda_1 = { 1\over 2} (-1-\sqrt{5})$, $\lambda_2 = { 1\over 2} (-1+\sqrt{5})$
and note that $(x-\lambda_1)(x-\lambda_2) = x^2+x-1$ and write
$f(x) = {1\over \sqrt{5}} ({1 \over x-\lambda_2} - {1 \over x-\lambda_1})$, expand
as a series (using something like ${1 \over 1-x} = 1+x+x^2+...$) and
compare coefficients to get the $a_n$.
| {
"language": "en",
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Without use of derivatives, prove that $e^{-x}\left(1+x+\frac{x^2}{2}\right)$ is strictly decreasing
Prove, without use of derivatives, that function $x\mapsto e^{-x}\left(1+x+\frac{x^2}{2}\right)$ is strictly decreasing.
Attempt. Functions $x\mapsto x^ne^{-x}$ for $n=0,1,2$ are not strictly decreasing (in order to guarantee that their sum is decreasing). Working separately on $e^{-x},~~1+x+\frac{x^2}{2}$ didn't work also, since $1+x+\frac{x^2}{2}$ is not decreasing on the reals. I am looking for an elementary way to prove it (it's derivative is $-\frac{e^{-x}x^2}{2}<0$ for $x\neq 0$).
Thanks in advance for the help.
| I will assume the properties $e^x > 1 +x + \frac{x^2}{2}$ and $e^{x+y} = e^x e^y$ are given.
For $x+h > x > 0$, we have $\left(h+ \frac{h^2}{2}\right)\left(1+x+ \frac{x^2}{2}\right)> h + xh+\frac{h^2}{2}$, and, hence,
$$e^h > 1+ h + \frac{h^2}{2} > 1 + \frac{h + xh+\frac{h^2}{2}}{1+x+ \frac{x^2}{2}} = \frac{1 + (x+h)+\frac{(x+h)^2}{2}}{1+x+ \frac{x^2}{2}}$$
Thus,
$$1+x+ \frac{x^2}{2} > e^{-h}\left(1 + (x+h)+\frac{(x+h)^2}{2} \right)$$
Multiplying both sides by $e^{-x}$ we get
$$e^{-x}\left(1+x+ \frac{x^2}{2}\right) > e^{-(x+h)}\left(1 + (x+h)+\frac{(x+h)^2}{2} \right)$$
Something similar should work for the case $x < 0$.
| {
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$\int_0^1 \frac{x^{p}}{x^{p+1}+(1-x)^{p+1}} dx=?$ $$\int_0^1 \frac{x^{p}}{x^{p+1}+(1-x)^{p+1}} dx=?$$
I tried to use
$$\int_0^1 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}} dx=\frac{1}{2}$$ and integration by parts. I do not know if there is any restriction on p.in original question p=2014,Question from Jalil Hajimir.
| Assume $p$ is a positive integer. Of course, you don't need it this strong.
First, observe that
\begin{align}
I(p) =& \int^1_0 \frac{x^p}{x^{p+1}+(1-x)^{p+1}}\ dx = \int^1_0 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{dx}{x}\\
=& \int^1_0 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{(1-x)}{x} dx+ \int^1_0 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ dx\ \ \ \ \ (1)\\
=&\ \int^1_0 \frac{(1-x)^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{x}{1-x} dx+ \int^1_0 \frac{(1-x)^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ dx\ \ \ \ \ (2)
\end{align}
which means
\begin{align}
2I(p) = \int^1_0 f(x)+f(1-x)\ dx +1 = 2\int^1_0 f(x)\ dx +1
\end{align}
where
\begin{align}
f(x) = \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{(1-x)}{x} = \frac{(\frac{1-x}{x})}{1+(\frac{1-x}{x})^{p+1}}.
\end{align}
Hence
\begin{align}
I(p) = \int^\infty_0 \frac{u}{(1+u^{p+1})(1+u)^2} du+\frac{1}{2}.
\end{align}
You can finish the rest using contour integration.
Edit: I realize I did a lot of unnecessary calculations. In fact, we have
\begin{align}
I(p) =& \int^1_0 \frac{1}{(1+(\frac{1-x}{x})^{p+1})}\frac{dx}{x} = \int^\infty_0 \frac{du}{(1+u^{p+1})(1+u)}
\end{align}
and
\begin{align}
\lim_{p\rightarrow \infty}I(p)=\lim_{p\rightarrow \infty}\int^\infty_0 \frac{du}{(1+u^{p+1})(1+u)} =\int^1_0 \frac{du}{1+u} = \ln(2).
\end{align}
| {
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Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$
Let's see what happens:-
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$
$$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$
In the same way for further terms, we will get $0$
Let's also confirm for general term
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$
$$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$
So the whole expression $E$ will be zero
But actual answer is $1$
Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
$OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$
Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?
I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.
But I got the following counter thought:-
$\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$
As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|<1$
$\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$
$\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$
Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity.
But surprisingly $\dfrac{1}{3}$ is the correct answer.
I am feeling very confused in these two things. Please help me.
| If you know harmonic numbers
$$S_n=\sum_{i=i}^n \frac 1 {n^2+i}=H_{n^2+n}-H_{n^2}$$ Using the asympotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
p^2}+O\left(\frac{1}{p^3}\right)$$ apply it twice and continue with Taylor series to get
$$n S_n=1-\frac{1}{2 n}-\frac{1}{6 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit nd how it is approached.
Moreover, this gives a quite good approximation of the sum. Using $n=10$, the exact value is $\frac{11210403701434961}{11818204429243212} \approx 0.94857$ while the above truncated series gives $\frac{569}{600}\approx 0.94833$.
| {
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"answer_count": 7,
"answer_id": 3
} |
evaluate $\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}$ $$
\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}
$$
I have tried the following:
$$
\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}=\\
\lim_{n\to\infty}\left(\frac{n^3}{n^3}\right)\cdot\left(\frac{1-\frac{1}{n}+\frac{2}{n^2}+\frac{3}{n^3}}{1+\frac{2}{n}-\frac{1}{n^3}}\right)^{\left(\frac{n^2}{n^2}\right)\left(\frac{8n-3+\frac{2}{n}-\frac{2}{n^2}}{10+\frac{3}{n}-\frac{1}{n^2}}\right)}$$
So I have a limit of $1^{\infty}$
| Just write
*
*$\frac{n^3-n^2+2n+3}{n^3+2n^2-1} = 1 - \underbrace{\frac{3n^2-2n-4}{n^3+2n^2-1}}_{a_n :=}$ and note that
*$\lim_{n\to \infty}(1-a_n)^{\frac 1{a_n}} = \frac 1e$
Now, you need only the limit
*
*$\lim_{n\to\infty}a_n\frac{8n^3-3n^2+2n-2}{10n^2+3n-1} = \lim_{n\to\infty}\frac{3n^2-2n-4}{n^3+2n^2-1}\frac{8n^3-3n^2+2n-2}{10n^2+3n-1} = \frac{12}{5}$
Putting all this together you get
$$\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)} = \lim_{n\to\infty}\left(\left(1-a_n\right)^{\frac{1}{a_n}}\right)^{a_n\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}$$
$$ = \left( \frac{1}{e}\right)^{\frac{12}{5}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show $x^2+y^2=9z+3$ has no integer solutions
Show $x^2+y^2=9z+3$ has no integer solutions
So I know that $x^2+y^2=3(3z+1)$
And since $3\mid (9z+3)$ and $3\not\mid (3z+1)$ then $9z+3$ cannot be a square since it has a prime divisor which has a power less then $2$.
So does know this isn't a pythagorean triple $(x,y,\sqrt{9z+3})$ give me that there are no solutions or do I have to show more?
| Since $x^2 + y^2 = 9z+3$ must be true, it must also be true that $$x^2+y^2 \equiv 3 \pmod 9$$
However, if we list out $x^2 \pmod 9$ for $0 \le x < 9$, we get $$0, 1, 4, 0, 7, 7, 0, 4, 1$$
There is no way of adding two numbers from $0, 1, 4, 7$ such that the sum is equal to $0 \pmod 9$.
Therefore, $x^2 + y^2 = 9z + 3$ has no integer solutions.
| {
"language": "en",
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"answer_id": 2
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Is this a decreasing sequence? In this nice question , the OP asks, or wants to prove that the sequence: $$a_{n+1} = \sqrt[n]{a_{1}+\dots+a_{n}}, \\a_1=1$$ is increasing. After playing a bit with Mathematica Cloud I realised that this is not the case as you may see in the following diagram:
Actually, it seems that it is decreasing for $n> 4$. I have tried to prove it with induction and some other basic methods but I find it hard....Any hints, suggestions, please?
Note that if it is decreasing there should be a limit and this limit should be $1$ as stated in these answers .
| For any integer $m \geqslant 3$,
\begin{align*}
\left(1 + \frac1m\right)^m & = 1 + \binom{m}1m^{-1} + \binom{m}2m^{-2} + \cdots + m^{-m+2} + m^{-m} \\
& \leqslant 1 + 1 + 1 + \cdots + \frac1m + \frac1m \quad (m + 1 \text{ terms}) \\
& < m.
\end{align*}
Therefore, $(m + 1)^m < m^{m+1}$. (I'm almost sure I once gave a shorter proof of this on MSE! $\ldots$ Apparently not. See the comments.)
That is, $n^{n-1} < (n - 1)^n,$ for every integer $n \geqslant 4.$
Equivalently,
$$
n^{\frac{n-1}n} < n - 1 \quad (n \geqslant 4).
$$
With $n$ any positive integer, let $f(x) = x^n - x^{n-1}.$ Then
$$
f'(x) = nx^{n-2}\left(x - 1 + \frac1n\right) > 0 \text{ if } x \geqslant 1,
$$
so $f(x)$ is strictly increasing for $x \geqslant 1.$
(That, too, must surely have a nicer proof. $\ldots$ Indeed it has. See the comments.)
Therefore, if $n \geqslant 4$ and $x \geqslant \sqrt[n]n,$
\begin{align*}
x^n - x^{n-1} & \geqslant n - n^{\frac{n-1}n} \\
& > 1, \\
\therefore\ x^{n+1} - x^n & > x.
\end{align*}
Take $x = a_{n+1} \geqslant \sqrt[n]n.$
Then $a_{n+2}^{n+1} = x^n + x < x^{n+1},$ therefore
$$
a_{n+2} < x = a_{n+1}.
$$
It remains to verify $a_5 < a_4.$
We have $43^2 = 1849 > 1800,$ therefore $43 > 30\sqrt2$, therefore
$$
2 + \sqrt2 < 45 - 29\sqrt2 = (3 - \sqrt2)^3,
$$
therefore $2 + \sqrt2 + \sqrt[3]{2 + \sqrt2} < 5,$ i.e. $a_5^4 < 5 < \frac{81}{16},$ therefore
$$
a_5 < \frac32.
$$
On the other hand, we have $11^2 = 121 < 128,$ therefore $11 < 8\sqrt2,$ therefore $2 + \sqrt2 > \frac{27}8,$ therefore
$$
a_4 > \frac32 > a_5.
$$
| {
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"source": "stackexchange",
"question_score": "2",
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value of $k$ in binomial sum
If $\displaystyle (1-x)^{\frac{1}{2}}=a_{0}+a_{1}x+a_{2}x^2+\cdots \cdots +\infty.$ and $a_{0}+a_{1}+a_{2}+\cdots+a_{10}=\frac{\binom{20}{10}}{k^{10}}.$ Then $k$ is
what i try
$(1-x)^{\frac{1}{2}}=1-\frac{1}{2}x+\frac{1}{2}\bigg(\frac{1}{2}-1\bigg)\frac{x^2}{2!}+\frac{1}{2}\bigg(\frac{1}{2}-1\bigg)\bigg(\frac{1}{2}-2\bigg)\frac{x^3}{3!}+\cdots\cdots$
$\displaystyle a_{0}=1,a_{1}=-\frac{1}{2},a_{2}=-\frac{1}{2^2}\frac{1}{2!},a_{3}=\frac{1\cdot 3}{2^3}\frac{1}{3!}+\cdots$
How do i solve it Help me pleses
| We have
$$\sum_{n=0}^{1}(-1)^n\binom{\frac 12}{n}=\frac 12=\frac 24=\frac{\binom{2}{1}}{4^1},\quad \sum_{n=0}^{2}(-1)^n\binom{\frac 12}{n}=\frac 38=\frac{6}{16}=\frac{\binom{4}{2}}{4^2}$$
$$\sum_{n=0}^{3}(-1)^n\binom{\frac 12}{n}=\frac{5}{16}=\frac{20}{64}=\frac{\binom{6}{3}}{4^3},\quad \sum_{n=0}^{4}(-1)^n\binom{\frac 12}{n}=\frac{35}{128}=\frac{70}{256}=\frac{\binom{8}{4}}{4^4}$$
So, it seems that the following holds :
$$\sum_{n=0}^{r}(-1)^n\binom{\frac 12}{n}=\frac{\binom{2r}{r}}{4^r}\tag1$$
One can prove $(1)$ by induction.
Proof :
$(1)$ holds for $r=0$.
Supposing that $(1)$ holds for $r$ gives
$$\begin{align}\sum_{n=0}^{r+1}(-1)^n\binom{\frac 12}{n}&=\sum_{n=0}^{r}(-1)^n\binom{\frac 12}{n}+(-1)^{r+1}\binom{\frac 12}{r+1}
\\\\&=\frac{\binom{2r}{r}}{4^r}+(-1)^{r+1}\cdot\frac{\frac 12(\frac 12-1)\cdots (\frac 12-r)}{(r+1)!}
\\\\&=\frac{\binom{2r}{r}}{4^r}+(-1)^{r+1}\cdot\frac{(2r-1)(2r-3)\cdots (2-1)(-1)}{(r+1)!(-2)^{r+1}}
\\\\&=\frac{\binom{2r}{r}}{4^r}+(-1)^{r+1}\cdot\frac{-(2r-1)(2r-3)\cdots (2-1)\cdot (2r)(2r-2)\cdots 2}{(r+1)!(-2)^{r+1}\cdot (2r)(2r-2)\cdots 2}
\\\\&=\frac{\binom{2r}{r}}{4^r}+(-1)^{r+1}\cdot\frac{-(2r)!}{(r+1)!(-2)^{r+1}\cdot 2^rr!}
\\\\&=\frac{\binom{2r}{r}}{4^r}+\frac{-1}{(r+1)\cdot 2^{2r+1}}\cdot\frac{(2r)!}{r!r!}
\\\\&=\frac{\binom{2r}{r}}{4^r}\bigg(1-\frac{1}{2(r+1)}\bigg)
\\\\&=\frac{1}{4^r}\cdot\frac{(2r)!}{r!r!}\cdot\frac{2r+1}{2r+2}\cdot\frac{2r+2}{2r+2}
\\\\&=\frac{\binom{2r+2}{r+1}}{4^{r+1}}\qquad\square\end{align}$$
It follows from $(1)$ that
$$a_0+a_1+\cdots +a_{10}=\sum_{n=0}^{10}(-1)^n\binom{\frac 12}{n}=\frac{\binom{20}{10}}{\color{red}4^{10}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding the area enclosed by the locus of the vertex of the rectangle at which the normals meet. Let a and b be the lengths of the semimajor and semiminor axes of an ellipse respectively.
Draw a rectangle whose two sides are tangent to the ellipse and the other two are normal to the ellipse.
I want to find the area enclosed by the locus of the vertex of the rectangle at which the normals meet.
Solution:-
My attempt:- We are required to find the locus of point $(h,k)$ from which two perpendicular lines can be drawn which are normal to the ellipse.
Normal to the ellipse at the point $(a \cos{\theta}, b\sin{\theta})$ is given by $ax\sec{\theta}-by\csc{\theta}=a^2-b^2$ and the slope of this normal is given by $m=\frac{a}{b}\tan{\theta}$
Now how to proceed futher?
Now by putting $x=h, y=k$ how to eliminate $\theta$ and write the equation in m?
And how to arrive at the final answer?
If any member knows the correct answer may reply with correct answer.
The graph of the vertex of the rectangle at which the normals meet provided to me is as follows
Answer provided to me for the required area is $(a-b)^2\pi$
I tried to plot the equation provided in the comment section, in www.wolframalpha.com but it failed. see Here
| Let $A(u,v)$ be the outer vertex of the rectangle which should lies on the director circle, hence
$$u^2+v^2=a^2+b^2$$
Refer to another answer of mine here, the mid-point of the polar (the chord $BD$) is
$$\frac{1}{\dfrac{u^2}{a^2}+\dfrac{v^2}{b^2}}
\begin{pmatrix}
u \\ v
\end{pmatrix}$$
Since diagonals bisect each others, the required locus for the inner vertex $C(x,y)$ is given by
$$
\begin{pmatrix}
x \\ y
\end{pmatrix}+
\begin{pmatrix}
u \\ v
\end{pmatrix}=
\frac{2}{\dfrac{u^2}{a^2}+\dfrac{v^2}{b^2}}
\begin{pmatrix}
u \\ v
\end{pmatrix}
$$
Therefore
$$
\begin{pmatrix}
x \\ y
\end{pmatrix}=
\left(
\frac{2}{\dfrac{u^2}{a^2}+\dfrac{v^2}{b^2}}-1
\right)
\begin{pmatrix}
u \\ v
\end{pmatrix}$$
Converting into polar coordinates:
$$r^2=(a^2+b^2)
\left(
\frac{1-\dfrac{b^2\cos^2 \theta}{a^2}-\dfrac{a^2\sin^2 \theta}{b^2}}
{1+\dfrac{b^2\cos^2 \theta}{a^2}+\dfrac{a^2\sin^2 \theta}{b^2}}
\right)^2$$
The area can be found by using Mathematica:
\begin{align}
& \quad \frac{1}{2} \int_0^{2\pi} r^2 d\theta \\
&=
2(a^2+b^2) \int_0^{\tfrac{\pi}{2}}
\left(
\frac{1-\dfrac{b^2\cos^2 \theta}{a^2}-\dfrac{a^2\sin^2 \theta}{b^2}}
{1+\dfrac{b^2\cos^2 \theta}{a^2}+\dfrac{a^2\sin^2 \theta}{b^2}}
\right)^2 d\theta \\
& = 2(a^2+b^2)
\left[
\theta-
\frac{2ab}{a^2+b^2} \tan^{-1} \left( \frac{a\tan \theta}{b} \right)+
\left( \frac{2ab}{a^2+b^2} \right)^2
\frac{(a^2-b^2)\sin 2\theta}{a^2+b^2-(a^2-b^2)\cos 2\theta}
\right]_0^{\tfrac{\pi}{2}} \\
& =
\pi(a-b)^2
\end{align}
Addendum
By substituting $r^2=x^2+y^2$ and $(\cos \theta,\sin \theta)=\dfrac{(x,y)}{r}$, we have an implicit equation
$$(a^2+b^2)(x^2+y^2)
\left(
\frac{x^2}{a^2}+\frac{y^2}{b^2}
\right)^2=
(a^2-b^2)^2
\left(
\frac{x^2}{a^2}-\frac{y^2}{b^2}
\right)^2$$
which is consistent to another forum here.
| {
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$\int_0^\infty \frac{\sin^n x}{x^m}dx$ could be expressed via $\pi$ or $\log$ I want to show some results first (they were computed by MMA)
$$
\int_0^\infty \frac{\sin^5 x}{x^3} dx =\frac{5}{32}{\color{Red}\pi} \quad
\int_0^\infty \frac{\sin^5 x}{x^5} dx =\frac{115}{384} {\color{Red}\pi} \\
\int_0^\infty \frac{\sin^5 x}{x^2} dx =\frac{5}{16}\,{\color{Red}\log}\, \frac{27}{5} \quad
\int_0^\infty \frac{\sin^5 x}{x^4} dx =-\frac{5}{96}(27\,{\color{Red }\log } \,3-25\,{\color{Red}\log }\,5) \\
$$
and
$$
\int_0^\infty \frac{\sin^6 x}{x^4} dx =\frac{1}{8} {\color{Red}\pi} \quad
\int_0^\infty \frac{\sin^6 x}{x^6} dx =\frac{11}{40} {\color{Red}\pi} \\
\int_0^\infty \frac{\sin^6 x}{x^3} dx =\frac{3}{16}\,{\color{Red}\log}\, \frac{256}{27} \quad
\int_0^\infty \frac{\sin^6 x}{x^5} dx ={\color{Red}\log}\, \frac{3^\frac{27}{16}}{4} \\
$$
As we can see, in the integral $\displaystyle\int_0^\infty \frac{\sin^nx}{x^m}dx$
, if $n-m$ is even, the integral is expressed via $\pi$, and if $n-m$ is odd, the integral is expressed via $\log$. Amazing for me, it seems always true such as
$$\int_0^\infty \frac{\sin^8 x}{x^2} dx =\frac{5\pi}{32}$$ and
$$\int_0^\infty \frac{\sin^8 x}{x^3} dx =\frac{9}{8}\log\frac{4}{3}$$
Do we have a general method to compute $\displaystyle\int_0^\infty \frac{\sin^nx}{x^m}dx$ which implies these laws?
My attempt
This post tells us how to compute $\displaystyle\int_0^\infty \frac{\sin^n x}{x^n}dx \tag{*}$
We can compute some other $\displaystyle\int_0^\infty \frac{\sin^nx}{x^m}dx$ via $(*)$, such as via the formulas
$$\displaystyle\int_{0}^{\infty}\dfrac{\sin^3 x}{x}\,dx = \dfrac{3}{4}\int_{0}^{\infty}\dfrac{\sin x}{x}\,dx - \dfrac{1}{4}\int_{0}^{\infty}\dfrac{\sin 3x}{x}\,dx$$
and
$$\int_0^\infty \frac{\sin^2 (2x)}{x^2}dx=\int_0^\infty \frac{4\sin^2 x-4\sin^4 x}{x^2}$$
But it's complicated to compute the general cases. Could you please share some ideas of a possible method to show the law mentioned above?
| If $n=2p$ is even then note that
$$\sin^n(x)=\frac1{2^n}\binom np+\frac1{2^{n-1}}\sum_{k=0}^{p-1}\binom nk(-1)^{p-k}\cos((n-2k)x)$$
If $n=2p+1$ is odd then note that
$$\sin^n(x)=\frac1{2^{n-1}}\sum_{k=0}^p\binom nk(-1)^{p-k}\sin((n-2k)x)$$
By rewriting the $\sin^n(x)$, we get an integral of a sum of $\cos(jx)/x^m$ or $\sin(jx)/x^m$. By repeatedly integrating by parts, one can reduce this down to an integrals of the forms:
$$\int_0^\infty\frac{\cos(ax)-\cos(bx)}x~\mathrm dx=\ln\frac ba,~\int_0^\infty\frac{\sin(ax)}x~\mathrm dx=\frac\pi2$$
From which you can easily deduce the result, depending on whether $n-m$ is even or odd, as either a multiple of $\pi$ or a multiple of a logarithm.
If $n-m$ is even, the computation can be reduced. Since $n-m$ is even, then the integrand is even and we can rewrite it as integral over $\mathbb R$:
$$\int_0^\infty\frac{\sin^n(x)}{x^m}~\mathrm dx=\frac12\int_{-\infty}^\infty\frac{\sin^n(x)}{x^m}~\mathrm dx$$
We can then apply the above formulas, take the real or imaginary parts, and then use a standard semicircle contour and complex analysis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Bounded sequence of reals Suppose that $\{a_{n} \}$ is a sequence of real numbers that satisfy $a_{i} + a_{j} \geq a_{i+j}$.
Then prove $$a_{n} \leq \sum_{i=1}^{n} \frac{a_{i}}{i}$$.
I tried to use straight-up induction, but I am getting stuck. What I have is:
$a_{n+1} \leq a_{1} + a_{n} \leq a_{1} + \sum_{i=1}^{n} \frac{a_{i}}{i}$, but I don't seem to be getting anywhere.
| From the inequality you easily obtain that for all integers $i_1, ..., i_k$, one has: $a_{i_1} + ... + a_{i_k} \geq a_{i_1 + ... + i_k}$. By taking them all equal, one gets: $a_{kl} \leq l \cdot a_k$, so $\frac{a_{kl}}{l} \leq a_k$.
We have:
$$
a_1 + a_{n-1} \geq a_n \\
... \\
a_{n-1} + a_{1} \geq a_n \\
$$
Summing all the inequalities gives $(n-1) \cdot a_n \leq 2\cdot(a_1 + \cdots + a_{n-1})$. Let us use this to prove the induction.
Assume the inequality is true for integers $k \leq n$. We will show it stands for $n+1$.
From what we proved we get $n \cdot a_{n+1} \leq 2\cdot(a_1 + \cdots + a_{n})$.
Thus,
$$
\begin{align*}
n \cdot a_{n+1}
&\leq 2\cdot \left( a_1 + (a_1 + \frac{a_2}{2}) + \cdots + (a_1 + \cdots + \frac{a_n}{n}) \right) \\
&\leq 2\cdot \left( n \cdot a_1 + (n-1) \cdot \frac{a_2}{2} + \cdots + 1 \cdot \frac{a_n}{n} \right) \\
\end{align*}
$$
Adding $a_{n+1}$ on both sides gives:
$$
\begin{align*}
(n+1) \cdot a_{n+1}
&\leq 2\cdot \left( n \cdot a_1 + (n-1) \cdot \frac{a_2}{2} + \cdots + 1 \cdot \frac{a_n}{n} \right) + a_{n+1} \\
\end{align*}
$$
But I do not know how to go further...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3479081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
For $a$, $b$, $c$ the sides of a triangle, show $ 7(a+b+c)^3-9(a+b+c)\left(a^2+b^2+c^2\right)-108abc\ge0$
If $a$, $b$, and $c$ are the three sidelengths of an arbitrary triangle, prove that the following inequality is true, with equality for equilateral triangles.
$$ 7\left(a+b+c\right)^3-9\left(a+b+c\right)\left(a^2+b^2+c^2\right)-108abc\ge0 \tag{1}$$
In expanded form:
$$ 6\left(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \right)-\left(a^3+b^3+c^3\right)-33abc\ge0 \tag{2}$$
This a part of an ongoing research in triangle geometry and related to solving a cubic equation.
| By AM-GM
$$\frac{a_1+\cdots+a_n}{n}\geq\sqrt[n]{a_1\cdots a_n}$$
Since $a,~ b,~ c$ are positive real numbers
$$\frac{a^3+b^3+c^3}{3}\geq\sqrt[3]{a^3b^3c^3}=abc$$
$$a^3+b^3+c^3\geq3abc$$
$$2(a^3+b^3+c^3)\geq6abc\tag{1}$$
Now we want to prove that
$$2a^2(b + c) + 2b^2(c + a) + 2c^2(a + b) ≥ a^3 + b^3 + c^3 + 9abc\tag{2}$$
First let
$$ \begin{cases}
a = y + z \\
b = z + x \\
c = x + y
\end{cases}
$$
With $x,~y,~z\geq0$, then the left side of $(2)$ becomes
$$4x^3 + 4y^3 + 4z^3 + 10x^2(y + z) + 10y^2(z + x) + 10z^2(x + y) + 24xyz$$
And the right side becomes
$$2x^3 + 2y^3 + 2z^3 + 12x^2(y + z) + 12y^2(z + x) + 12z^2(x + y) + 18xyz$$
Further simplify we have
$$x^3 + y^3 + z^3 + 3xyz ≥ x^2(y + z) + y^2(z +x) + z^2(x + y)$$
which is Schur's inequality, so we have proved that $(2)$ holds true.
From $(2)$ we have
$$6(a^2(b + c) + b^2(c + a) + c^2(a + b)) ≥ 3(a^3 + b^3 + c^3) + 27abc\tag{3}$$
Add $(1)$ and $(3)$
$$6\left(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \right)+2(a^3+b^3+c^3)\ge 3(a^3+b^3+c^3)+33abc$$
$$ 6\left(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \right)-\left(a^3+b^3+c^3\right)-33abc\ge0 $$
and we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Trying to find $\cot^{-1}x-\cot^{-1}y$ for $\forall$ $x,y$ Suppose we want to calculate $\cot^{-1}x-\cot^{-1}y$ for $\forall~x,y$
$$\cot^{-1}x-\cot^{-1}y=\theta\tag{1}$$
Let's find range of $\theta$, assuming $x$ and $y$ to be independent variables
$$\theta\in(-\pi,\pi)$$
Taking $\cot$ on both sides of equation $1$
$$\dfrac{\cot(\cot^{-1}x)\cot(\cot^{-1}y)+1}{\cot(\cot^{-1}y)-\cot(\cot^{-1}x)}=\cot\theta$$
$$\dfrac{xy+1}{y-x}=\cot\theta$$
Taking $\cot^{-1}$ on both sides
$$\cot^{-1}\dfrac{xy+1}{y-x}=\cot^{-1}(\cot\theta)$$
$$\cot^{-1}(\cot\theta)=\begin{cases}
\pi+\theta,&-\pi<\theta<0 \\
\theta,&0<\theta<\pi \\
\end{cases}$$
So
$$\theta=\begin{cases}
-\pi+\cot^{-1}\dfrac{xy+1}{y-x},&-\pi<\theta<0 \\
\cot^{-1}\dfrac{xy+1}{y-x},&0<\theta<\pi \\
\end{cases}$$
$$\cot^{-1}x-\cot^{-1}y=\begin{cases}
-\pi+\cot^{-1}\dfrac{xy+1}{y-x},&-\pi<\cot^{-1}x-\cot^{-1}y<0 \\
\cot^{-1}\dfrac{xy+1}{y-x},&0<\cot^{-1}x-\cot^{-1}y<\pi\\
\end{cases}$$
$$\cot^{-1}x-\cot^{-1}y=\begin{cases}
-\pi+\cot^{-1}\dfrac{xy+1}{y-x},&\cot^{-1}x-\cot^{-1}y\in(-\pi,0) \\
\cot^{-1}\dfrac{xy+1}{y-x},&\cot^{-1}x-\cot^{-1}y\in(0,\pi)\\
\end{cases}$$
$$\cot^{-1}x-\cot^{-1}y=\begin{cases}
-\pi+\cot^{-1}\dfrac{xy+1}{y-x},&\dfrac{xy+1}{y-x}\in(-\infty,\infty) \\
\cot^{-1}\dfrac{xy+1}{y-x},&\dfrac{xy+1}{y-x}\in(-\infty,\infty)\\
\end{cases}$$
So finally we can write
$$\cot^{-1}x-\cot^{-1}y=\begin{cases}
-\pi+\cot^{-1}\dfrac{xy+1}{y-x}, &x>y\\
\cot^{-1}\dfrac{xy+1}{y-x}, &x<y\\
\end{cases}$$
We can also derive for $\cot^{-1}x+\cot^{-1}y$ in the following way
Adding $\pi$ to both sides
$$\cot^{-1}x+\pi-\cot^{-1}y=\begin{cases}
\cot^{-1}\dfrac{xy+1}{y-x}, &x>y\\
\pi+\cot^{-1}\dfrac{xy+1}{y-x}, &x<y\\
\end{cases}$$
$$\cot^{-1}x+\cot^{-1}(-y)=\begin{cases}
\cot^{-1}\dfrac{xy+1}{y-x}, &x>y\\
\pi+\cot^{-1}\dfrac{xy+1}{y-x}, &x<y\\
\end{cases}$$
Replacing $y$ with $-y$
$$\cot^{-1}x+\cot^{-1}(y)=\begin{cases}
\cot^{-1}\dfrac{-xy+1}{-y-x}, &x>-y\\
\pi+\cot^{-1}\dfrac{-xy+1}{-y-x}, &x<-y\\
\end{cases}$$
$$\cot^{-1}x+\cot^{-1}(y)=\begin{cases}
\cot^{-1}\dfrac{xy-1}{y+x}, &x+y>0\\
\pi+\cot^{-1}\dfrac{xy-1}{y+x}, &x+y<0\\
\end{cases}$$
Is it correct, I am asking because I am not sure about it because in text-books I just find $\cot^{-1}x-\cot^{-1}y=\cot^{-1}\dfrac{xy+1}{y-x}$ and I was not finding formula for $\cot^{-1}x+\cot^{-1}(y)$.
| $$0<\cot^{-1}x+\cot^{-1}y<2\pi$$
So, $\cot^{-1}x+\cot^{-1}y$ will be $\cot^{-1}\dfrac{xy-1}{x+y}$ if $\cot^{-1}x+\cot^{-1}y<\pi$
$\iff\tan^{-1}x+\tan^{-1}y>0\iff\tan^{-1}x>-\tan^{-1}y=\tan^{-1}(-y)$
As $\tan^{-1}x$ is strictly increasing in $\left(-\dfrac\pi2,\dfrac\pi2\right),$ we need $x>-y\iff x+y>0$
So, $\cot^{-1}x+\cot^{-1}y$ will be $\pi+\cot^{-1}\dfrac{xy-1}{x+y}$ if $\pi<\cot^{-1}x+\cot^{-1}y<2\pi$
$\iff-\pi<\tan^{-1}x+\tan^{-1}y<0\iff x+y<0$
Now replace $y$ with $-z$
and use $\cot^{-1}(-z)=\pi-\cot^{-1}z$ like How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3493923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$
I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$ but got stucked, and made it more complex, any help?
Sorry if I made any silly mistake, it's been while since I practiced complex equation and finding roots. Was helping my brother with his doubts :)
| $$
(x^2-1)(x^4+x^2+1) = x^6 - 1
$$
$x^6-1=0$ if and only if $x$ is one of the $6$th roots of $1.$
$(x^2-1)(x^4+x^2+1)=0$ if and only if either $x^2-1=0,$ in which case $x=\pm 1,$ or $x^4+x^2+1=0,$ which fails to hold if $x=\pm 1,$ since $(\pm 1)^4+(\pm 1)^2 + 1 = 3\ne0.$ Thus the roots of $x^4+x^2+1=0$ are the $6$th roots of $1$ other than $\pm 1.$ Their $6$th powers are equal to $1,$ so the sum of their $6$th powers is $4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
Limit of sum $x^3+x^5+x^7+x^9+...)$ I am asked to give the limit of:
$$ x^3+x^5+x^7+x^9+... \quad x\in(-1,1)$$
So I do the following:
The sum of the first $n$ terms will be equal to:
$$x^3+x^5+x^7+...+x^{3+2(n-1)}$$
I factor out $x^3$, I get:
$$x^3(1+x^2+x^4+..+x^{2(n-1)})$$
I also factor out $x^2$, I get:
$$x^3 x^2(1/x^2+1+x^2+x^3+...+x^{n-1}=x^5(1/x^2+1+x^2+x^3+...+x^{n-1})$$
So I have:
$$x^5/x^2+x^5(1+x^2+x^3+...+x^{n-1})=x^3+\frac{x^5 (1-x^n)}{1-x}$$
So I take the limit when $n$ goes to infinity and I get:
$$x^3+\frac{x^5}{1-x}=\frac{x^3(1-x^2)}{1-x^2}+\frac{(1+x)x^5}{1-x^2}=\frac{x^3-x^5+x^5+x^6}{1-x^2}=\frac{x^3+x^6}{1-x^2}$$
But the right answer is $\frac{x^3}{1-x^2}$
Where did I go wrong?
| Note that
$$1+x^2+x^4+\dots =\sum_{n=0}^\infty x^{2n}=\sum_{n=0}^\infty (x^2)^n$$
is convergent for $x\in (-1,1)$. Thus, call
$$S(x)=1+x^2+x^4+\dots$$
Then we have
$$S(x)=1+x^2\left(1+x^2+x^4+\dots\right)=1+x^2S(x)$$
Solving for $S(x)$ gives us
$$S(x)=\frac{1}{1-x^2}$$
(you could also get $S(x)$ by noting that it is a geometric series in $x^2$). Now, we have the relation that
$$x^3+x^5+x^7+\dots =x^3(1+x^2+x^4+\dots)=x^3 S(x)$$
Thus
$$x^3+x^5+x^7+\dots=\frac{x^3}{1-x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving: $\int\cos^4(x)\sin^4(x)\ dx$ I've been trying to solve this problem - $\int\cos^4(x)\sin^4(x)\ dx$, but I don't seem to be succeeding. This is what I've done: $$\int\left(\frac{1+\cos(2x)}{2}\right)^2\left(\frac{1-\cos(2x)}{2}\right)^2\ dx\\=\frac1{16}\int\left(1+2\cos(2x)+\cos^2(2x)\right)\left(1-2\cos(2x)+\cos^2(2x)\right)\ dx\\=\frac1{16}\int1+2\cos^2(2x)+\cos^4(2x)\ dx\\=\frac1{16}\int1+1+\cos(4x)+\frac{
(1+\cos(4x))^2}{4}\ dx\\=\frac1{16}\int2+\cos(4x)+\frac{1+2\cos(4x)+\cos^2(4x)}{4}\ dx\\=\frac1{16}\int2+\cos(4x)+\frac{1}{4}+\frac{\cos(2x)}{2}+\frac{1+\cos(8x)}{8}\ dx\\=\frac1{16}\left(2x+\frac{\sin(4x)}{4}+\frac{x}4+\frac{\sin(2x)}{4}+\frac{x}8+\frac{\sin(8x)}{64}\right)\\=\frac{x}{8}+\frac{\sin(4x)}{64}+\frac{x}{64}+\frac{\sin(2x)}{64}+\frac{x}{128}+\frac{\sin(8x)}{1024}\\=\frac{19x}{128}+\frac{\sin(2x)}{64}+\frac{\sin(4x)}{64}+\frac{\sin(8x)}{1024}$$
I checked this up on integral-calculator.com. They seem to have a similiar answer - but not quite the same - their denominators are different and they don't have a $\sin(2x)$ term. I keep retrying this problem, but I seem to be missing something. Where am I going wrong?
| Hint: An alternative approach is to use the identity
$$\sin(2x) = 2\sin(x)\cos(x)$$
Then we get that
\begin{align*}
\int \cos^4(x)\sin^4(x)dx &= \int \big(\cos(x)\sin(x)\big)^4 dx\\
&= \int \left(\frac{\sin(2x)}{2}\right)^4 dx \\
&= \frac{1}{16} \int \sin^2(2x) \sin^2(2x)dx \\
&= \frac{1}{16} \int \left(\frac{1-\cos(4x)}{2}\right)\left(\frac{1-\cos(4x)}{2}\right)dx
\end{align*}
And then the rest I leave as an exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Reducing $\frac{\left(\left(\sqrt{\frac{a + b}{a - b}}\right)^2+1\right)\cdot c}{2c\sqrt{\frac{a + b}{a - b}}}$ to $\frac{1}{\sqrt{1 - b^2/a^2}} $ I want to rearrange
$$ \frac{\left(\left(\sqrt{\frac{a + b}{a - b}}\right)^2+1\right) \cdot c}{2c \cdot \sqrt{\frac{a + b}{a - b}}} $$
into $$ \frac{1}{\sqrt{1 - b^2/a^2}} $$
I would be grateful for some hints or full explanation.
I do know you can get rid of $c$ and the root above easily.
| HINTS: After cancelling the $c$ as you mention, we have
\begin{align}
\frac{\left(\sqrt{\frac{a + b}{a - b}}\right)^2+1}{2 \sqrt{\frac{a + b}{a - b}}}
= \frac{\left(\sqrt{\frac{a + b}{a - b}}\right)^2}{2 \sqrt{\frac{a + b}{a - b}}} + \frac{1}{{2 \sqrt{\frac{a + b}{a - b}}}}
\end{align}
Now you can easily cancel stuff in the first term. Do you see how?
If you get stuck along the way, check out the spoilers by hovering your mouse over them (if you're on a computer).
Step 1: Canceling as mentioned above.
$$=\frac{\sqrt{\frac{a + b}{a - b}}}{2} + \frac{1}{{2 \sqrt{\frac{a + b}{a - b}}}} = \frac{\sqrt{a + b}}{2\sqrt{{a - b}}} + \frac{\sqrt{a - b}}{2\sqrt{a + b}}$$
Step 2: Make denominators equal.
$$=\frac{a + b}{2\sqrt{{(a - b)(a + b)}}} + \frac{a - b}{2\sqrt{(a + b)(a - b)}}$$
Step 3: Now add and cancel.
$$=\frac{a + b + (a - b)}{2\sqrt{{(a - b)(a + b)}}} = \frac{2a}{2\sqrt{{(a - b)(a + b)}}}$$
Step 4: Bring everything under the square root.
$$= \frac{a}{\sqrt{{a^2 - b^2}}} = \sqrt{\frac{a^2}{a^2 - b^2}}$$
Step 5: Divide to get the desired form.
$$ = \sqrt{\frac{1}{\frac{a^2}{a^2} - \frac{b^2}{a^2}}} = \frac{1}{\sqrt{1 - b^2 / a^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\begin{cases} x^4+x^2y^2+y^4=21 \\ x^2+xy+y^2=3 \end{cases}$
I should solve the following system: $$\begin{cases} x^4+x^2y^2+y^4=21 \\
x^2+xy+y^2=3 \end{cases}$$
by reducing the system to a system of second degree.
What can I look for in such situations? What is the way to solve this kind of systems? The only thing I see here is that we can factor:
$$\begin{cases} x^2(x^2+y^2)+y^4=21 \\
x(x+y)+y^2=3 \end{cases}$$
| I found $$9=(x^2+xy+y^2)^2=$$ $$=(x^4+x^2y^2+y^4)+2(x^2y^2+x^3y+xy^3)=$$ $$=21+2xy(xy+x^2+y^2)=21+2xy(3)=$$ $$=21+6xy$$ giving $xy=-2.$
Then $(x+y)^2=(x^2+xy+y^2)+xy=3+xy=3-2=1.$
Knowing $(S,P)=(\pm 1,-2),$ we can find all possible $\{x,y\}$ as the solutions of $0=z^2-Sz+P=0=z^2\pm z-2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Sum of the reciprocal $P$-smooth numbers for $P = \{ p, q \}$ ?? For distinct primes $p$ and $q$ (where $p < q$), the following sum
$$\sum_{n=0}^{\infty} \sum_{k=0}^n \frac{1}{p^n q^k}$$
yields the expression $\frac{f}{g}$ such that
$$f = p^2 q$$
and
$$g \equiv 1 \ (\text{mod} \ p).$$
The problem I'm having is finding an expression for $g$ in terms of $p$ and $q$. I'm probably just overlooking something.
| Wikipedia's Geometric series article gives the formulas for the sums of a finite & infinite geometric series of
$$\sum_{k = 0}^{n-1}ar^k = a\left(\frac{1-r^n}{1-r}\right) \tag{1}\label{eq1A}$$
$$\sum_{k = 0}^{\infty}ar^k = \frac{a}{1-r}, \text{ for } \left|r\right| \lt 1 \tag{2}\label{eq2A}$$
Using \eqref{eq1A} and \eqref{eq2A}, plus making a few other manipulations as shown below, gives the following simplifications of your double summation
$$\begin{equation}\begin{aligned}
\sum_{n=0}^{\infty} \sum_{k=0}^n \frac{1}{p^n q^k} & = \sum_{n=0}^{\infty}\frac{1}{p^n} \sum_{k=0}^n \frac{1}{q^k} \\
& = \sum_{n=0}^{\infty}\frac{1}{p^n}\left(\frac{1-\left(\frac{1}{q}\right)^{n+1}}{1-\frac{1}{q}}\right) \\
& = \sum_{n=0}^{\infty}\frac{1}{p^n}\left(\frac{q-\left(\frac{1}{q}\right)^{n}}{q - 1}\right) \\
& = \left(\frac{1}{q-1}\right)\left(q\sum_{n=0}^{\infty}\frac{1}{p^n} - \sum_{n=0}^{\infty}\left(\frac{1}{pq}\right)^n \right) \\
& = \left(\frac{1}{q-1}\right)\left(\frac{q}{1-\frac{1}{p}} - \frac{1}{1 - \frac{1}{pq}}\right) \\
& = \left(\frac{1}{q-1}\right)\left(\frac{pq}{p - 1} - \frac{pq}{pq - 1}\right) \\
& = \left(\frac{pq}{q-1}\right)\left(\frac{(pq - 1) - (p - 1)}{(p - 1)(pq - 1)}\right) \\
& = \left(\frac{pq}{q-1}\right)\left(\frac{pq - p}{(p - 1)(pq - 1)}\right) \\
& = \left(\frac{p^2q}{q-1}\right)\left(\frac{q - 1}{(p - 1)(pq - 1)}\right) \\
& = \frac{p^2q}{(p - 1)(pq - 1)}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
As such, you have
$$g = (p - 1)(pq - 1) \tag{4}\label{eq4A}$$
Note this also confirms the modulo statement you have of
$$g \equiv 1 \pmod p \tag{5}\label{eq5A}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
What is $2018^{2018}$ mod $20$? The answer is $4$.
I saw someone answered in $5$ seconds. How did they do that?
| Since $2018 \equiv -2 \pmod{20}$ we have $$2018^{2018} \equiv (-2)^{2018} \equiv 4^{1009}\pmod{20}$$
and then
$$4^1 \equiv 4 \pmod{20} \\ 4^2 = 16 \equiv -4 \pmod{20} \\ 4^3 \equiv -4\cdot 4 \equiv 4 \pmod{20} \\ \vdots$$
Since $1009$ is odd and the powers of $4$ alternate $4, -4, 4, -4, \ldots$, the answer is $4.$
| {
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Solving logarithm leaving in terms of $p$ and $q$ I would like to check the steps if Part a) is done correctly.
For Part b), how do I continue from below? I seem to stuck for $\log_{10}(5)$…
Here is the problem:
Given that $p = \log_{10} 2$ and $q = \log_{10} 7$, express the following in terms of $p$ and $q$.
a) $\log_{7} 4 = \frac{\log_{10} 4}{\log_{10} 7} = \frac{2 \log_{10} 2}{\log_{10} 7} = \frac{2p}{q}$
b) $\log_{10} \sqrt[3]{\frac{25}{49}} = \log_{10}5^\frac{2}{3} - \log_{10}7^\frac{2}{3} = \frac{2}{3}\log_{10}5 - \frac{2q}{3}$
[Source]
| Part a) is perfect (nice job)
For part b) the $\log_{10} 5$ is throwing a monkey wrench and you can't express $5$ in terms of $2$ and $7$ using only multiplication and division and exponents.
BUt you have two other values at your disposal: $\log_{10} 10 = 1$ and $\log_{10} 1 = 0$.
Can you express $5$ in terms of $2, 7, 10, 1$ in terms of multiplication and division and exponents. (Hint: $1$ is fairly useless as it is the multiplicative identity)
Hint:
$5 = \frac {10}2$
Solution:
$\log_{10} 5 = \log_{10} \frac {10}2 = \log_{10} 10 - \log_{10} 2 = 1 - p$ so $\log_{10} \sqrt[3]{\frac {25}{49}}=\frac 23\log_{10} 5-\frac {2q}3 = \frac 23(1-p)-\frac {2q}3$
Alternatively:
$\log_{10} \sqrt[3]{\frac {25}{49}}=\frac 23\log \frac 57=\frac 23\log \frac {10}{2*7} = \frac 23(\log 10 - (\log 2 + \log 7)) = \frac 23(1-p-q)$
| {
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Prove no two numbers besides 1 in the sequences are identical Question:
Given that {$a_n$} and {$b_n$} are two sequences of integers defined by
$$a_1=1,a_2=10,a_{n+1}=2a_n+3a_{n-1}$$
$$b_1=1,b_2=8,b_{n+1}=3b_n+4b_{n-1}$$
for $n=2,3,4,...$
Prove that, besides the number '1' , no two numbers in the sequences are identical.
What have I done so far
I have tried prove by contradiction, parity and the usual way to solve recursive problem but failed, does anyone can tell me some hints on how to approach this problem?
| Using the method of characteristic equation and roots, given in Linear difference equation,
gives for the first sequence the characteristic equation of
$$\begin{equation}\begin{aligned}
\lambda^{2} & = 2\lambda + 3 \\
\lambda^{2} - 2\lambda - 3 & = 0 \\
(\lambda - 3)(\lambda + 1) & = 0
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus, $\lambda = 3, -1$, giving the equation of
$$a_t = c_1(3)^{t} + c_2(-1)^{t} \tag{2}\label{eq2A}$$
Using $a_1 = 1$ and $a_2 = 10$ gives
$$1 = 3c_1 - c_2 \tag{3}\label{eq3A}$$
$$10 = 9c_1 + c_2 \tag{4}\label{eq4A}$$
Adding these $2$ equations gives $11 = 12c_1 \implies c_1 = \frac{11}{12}$. Using this in \eqref{eq3A} gives $c_2 = 3c_1 - 1 = \frac{21}{12}$. Thus, you have for all $n \ge 1$ that
$$a_n = \frac{11(3^{n}) + 21(-1)^{n}}{12} \tag{5}\label{eq5A}$$
For the second sequence, using a similar procedure (I'll leave this up to you to do), you will get for all $n \ge 1$ that
$$b_n = \frac{9(4^{n}) + 16(-1)^{n}}{20} \tag{6}\label{eq6A}$$
Since both sequences are strictly increasing, no $2$ values among the same sequence can be equal to each other. Instead, for some $i \gt 1$ and $j \gt 1$, assume $a_i = b_j$. You then get
$$\begin{equation}\begin{aligned}
\frac{11(3^{i}) + 21(-1)^{i}}{12} & = \frac{9(4^{j}) + 16(-1)^{j}}{20} \\
\frac{11(3^{i-1}) + 7(-1)^{i}}{4} & = \frac{9(4^{j-1}) + 4(-1)^{j}}{5} \\
5(11(3^{i-1}) + 7(-1)^{i}) & = 4(9(4^{j-1}) + 4(-1)^{j})
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Since $a_2 = 10$ and all $b_j \gt 10$ for $j \ge 2$, then consider $i \gt 2$. Then $9 \mid 3^{i-1}$, giving that \eqref{eq7A} becomes, modulo $9$,
$$\begin{equation}\begin{aligned}
5(11(3^{i-1}) + 7(-1)^{i}) & \equiv 4(9(4^{j-1}) + 4(-1)^{j}) \pmod 9 \\
35(-1)^{i} & \equiv 16(-1)^{j} \pmod 9 \\
(-1)(-1)^{i} & \equiv (-2)(-1)^{j} \pmod 9 \\
(-1)^{i+1} & \equiv (-2)(-1)^{j} \pmod 9
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
The LHS is either $-1$ or $1$ while the RHS is either $-2$ or $2$, so they are never equal to each other. This shows there are never $2$ values, apart from $a_1 = b_1 = 1$, of the sequences which are equal to each other.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3505751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Set of vectors defined by an equation Let $W$ be a set of vectors given by an equation $2x_1 + x_2 - 2x_3 + 3x_4 = 5$. I need to write this into a affine subspace form like $W = p + span(s)$ and I have absolutely no idea how should I proceed with this kind of definition.
| Note that
\begin{align}
\textsf{W} &= \left\{
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \in \mathbb{R}^4 :\, 2x_1 + x_2 - 2x_3 + 3x_4 = 5
\right\} \\
&= \left\{
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \in \mathbb{R}^4 :\, x_2 = 5 - 2x_1 - 2x_3 - 3x_4
\right\} \\
&= \left\{
\begin{pmatrix} x_1 \\ 5 - 2x_1 + 2x_3 - 3x_4 \\ x_3 \\ x_4 \end{pmatrix} :\, x_1,x_3,x_4 \in \mathbb{R}
\right\} \\
&= \left\{
\begin{pmatrix} 0 \\ 5 \\ 0 \\ 0 \end{pmatrix} + x_1\begin{pmatrix} 1 \\ -2 \\ 0 \\ 0 \end{pmatrix} + x_3\begin{pmatrix} 0 \\ 2 \\ 1 \\ 0 \end{pmatrix} + x_4\begin{pmatrix} 0 \\ -3 \\ 0 \\ 1 \end{pmatrix} :\, t_1,t_2,t_3 \in \mathbb{R}
\right\}
\end{align}
so, putting
$$p = \begin{pmatrix} 0 \\ 5 \\ 0 \\ 0 \end{pmatrix} \textrm{ and } S = \left\{ \begin{pmatrix} 1 \\ -2 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ -3 \\ 0 \\ 1 \end {pmatrix} \right\}$$
we have $\textsf{W} = p + \operatorname{span}(S)$.
| {
"language": "en",
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Problem in deducing an identity related to Logarithm to the base e . I have a problem in deriving this inequality about the logarithm to base $e$ .
Can someone please give hints.
If $x>1$ then prove that $$\log x + \log\frac{x}2+ \log \frac {x} {2^2} + \log \frac{x} {2^3} +\dots < \log^2 (x) .$$
Can someone please tell how to prove this?
| Because
$$\begin{align}
&\frac{\ln^2x}{\ln x+\ln\frac x2+\ln\frac x{2^2}+\cdots}\\
= &\frac{\ln^2x}{\ln\left(\displaystyle\prod_{i=0}^n\frac x{2^i}\right)}\\
= &\frac{\ln^2x}{\ln\left(\frac{x^n}{2^\frac{n(n-1)}2}\right)}\\
= &\frac{\ln^2x}{n\ln x-\frac{n^2-n}2\ln 2}\\
= &\frac{1}{\frac n{\ln x}-\frac{n^2-n}2\cdot\frac{\ln 2}{\ln^2x}}\\
= &\frac{1}{-\frac{\ln 2}{2\ln^2x}n^2+\left(\frac 1{\ln x}+\frac{\ln 2}{2\ln^2x}\right)n}
\end{align}$$
$\because\forall x>1,\ln x>0,\therefore-\frac{\ln 2}{2\ln^2x} <0,\therefore \lim\limits_{n\rightarrow\infty}-\frac{\ln 2}{2\ln^2x}n^2+\left(\frac 1{\ln x}+\frac{\ln 2}{2\ln^2x}\right)n =-\infty$.
$$\begin{align}
\therefore & \,\lim_{n\rightarrow\infty}\frac{1}{-\frac{\ln 2}{2\ln^2x}n^2+\left(\frac 1{\ln x}+\frac{\ln 2}{2\ln^2x}\right)n} = \frac 1{-\infty}=0^-\\
\therefore & \,\frac{\ln^2x}{\ln x+\ln\frac x2+\ln\frac x{2^2}+\cdots}<0\\
\text{and}\,\because & \,\forall x>1,\ln^2x>0,\therefore \ln x+\ln\frac x2+\ln\frac x{2^2}+\cdots <0<\ln^2x
\end{align}$$
So $\ln x+\ln\frac x2+\ln\frac x{2^2}+\cdots<\ln^2x$
| {
"language": "en",
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"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_0^\pi\frac{\sin\left(n+\frac{1}{2}\right)x}{\sin \frac{x}{2}}dx$
Evaluate $$
\int_0^\pi\frac{\sin\Big(n+\frac{1}{2}\Big)x}{\sin \frac{x}{2}}dx
$$
$$
\int_0^\pi\frac{\sin\Big(n+\frac{1}{2}\Big)x}{\sin \frac{x}{2}}dx=\int_0^\pi\frac{\sin\Big(nx+\frac{x}{2}\Big)}{\sin \frac{x}{2}}dx=\int_0^\pi\frac{\sin nx.\cos\frac{x}{2}+\cos nx.\sin\frac{x}{2}}{\sin\frac{x}{2}}dx\\
=\int_0^\pi\sin nx.\cot\frac{x}{2}.dx+\int_0^\pi\cos nx.dx\\
$$
I don't think it is leading anywhere, anyone could possibly help with how to approach this definite integral ?
Note: The solution given in my reference is $\pi$
| Note that
$$2\sin\frac x2\cos x = \sin\frac32x -\sin\frac12x$$
$$2\sin\frac x2\cos 2x = \sin\frac52x -\sin\frac32x$$
$$…$$
$$2\sin\frac x2\cos nx =\sin(n+\frac12)x -\sin(n-\frac12)x $$
Sum up both sides,
$$2\sin\frac x2 (\cos x + \cos 2x + … +\cos nx )= \sin(n+\frac12)x - \sin\frac x2 $$
Therefore,
$$
\int_0^\pi\frac{\sin\left(n+\frac{1}{2}\right)x}{\sin \frac{x}{2}}dx$$
$$=2\int_0^{\pi}(\cos x + \cos 2x + … +\cos nx )dx+\int_0^{\pi}dx= \pi$$
where all the cosine integrals vanish.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
if $mx+3|x+4|-2=0$ has no solutions, solve for $m$
If $mx+3|x+4|-2=0$ has no solutions, which of the following value could be $m$?
(A)5
(B)$-\frac{1}{2}$
(C)-3
(D)-6
(E)$\frac{10}{3}$
my attempt:
$$mx-2=-3|x+4| \\ m^2x^2-4mx+4=9x^2+72x+144 \\ (9-m^2)x^2+(4m+72)x+140=0$$
because the equation has no solutions, therefore
$$(4m+72)^2-4(9-m^2)140<0 \\ 576m^2+576m+144<0 \\ 4m^2+4m+1<0 \\ (2m+1)^2<0$$
maybe I made a mistake but I couldn't find it
| Consider the fact that $|x+4|$ has 2 different possible values. For $x\geq-4$, it becomes $x+4$. For $x<-4$, it becomes $-x-4$. Divide the problem into 2 those two different cases.
For $x\geq-4$, the equation becomes $mx+3(x+4)-2=mx+3x+10=(m+3)x+10=0$. We obtain that $x=-\frac{10}{m+3}$. In order to make there exist no solution for $x$, the value of denominator shall be zero. Thus $m+3=0$ which implies that $m=-3$.
For $x<-4$, the equation becomes $mx+3(-x-4)-2=mx-3x-14=(m-3)x-14=0$. We obtain that $x=-\frac{14}{m-3}$. Similarly, in order to make there exist no solution for $x$, the value of denominator shall be zero. Thus $m-3=0$ which implies that $m=3$.
After getting two different values of $m$ from both cases, we shall check whether it is correct or not by inserting each value of $m$ and try to solve the equation. After checking the value of $m$, the only value that fulfill the requirement is $m=-3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Stirling Numbers of the Second Kind Proof
Prove that
\begin{align*}
\sum_{n=1}^\infty S(n,n-2)x^n=\dfrac{x^3(1+2x)}{(1-x)^5}
\end{align*}
My guess is that I have to take the LHS and simply it, as well as take the RHS and simplify it, but not sure how to exactly do that.
Any help, tips, or a fully worked out solution would be appreciated!
|
Answer $1$ : Using the recurrence relation.
The Stirling numbers of the second kind satisfy the recurrence relation
\begin{eqnarray*}
{n \brace k}= k {n-1 \brace k}+ {n-1 \brace k-1}.
\end{eqnarray*}
With $k=n$ we have ${n \brace n}= {n-1 \brace n-1}=1$ and
\begin{eqnarray*}
S_{0}(x)=\sum_{n=1}^{\infty} {n \brace n} x^n = \frac{x}{1-x}. \\
\end{eqnarray*}
For $k=n-1$ we have
\begin{eqnarray*}
S_{1}(x)=\sum_{n=2}^{\infty} {n \brace n-1} x^n &=& \sum_{n=2}^{\infty} (n-1){n-1 \brace n-1} x^n+ \sum_{n=2}^{\infty} {n-1 \brace n-2} x^n\\
&=& x^2 \frac{d}{dx} \left( \frac{S_0(x)}{x} \right) + x S_1(x) \\
\end{eqnarray*}
Differentiating & rearranging gives
\begin{eqnarray*}
S_{1}(x)=\frac{x^2}{(1-x)^3}. \\
\end{eqnarray*}
For $k=n-2$ we have
\begin{eqnarray*}
S_{2}(x)=\sum_{n=3}^{\infty} {n \brace n-2} x^n &=& \sum_{n=3}^{\infty} (n-1){n-1 \brace n-2} x^n+ \sum_{n=3}^{\infty} {n-1 \brace n-3} x^n\\
&=& x^3 \frac{d}{dx} \left( \frac{S_1(x)}{x} \right) + x S_2(x) \\
\end{eqnarray*}
Again differentiating & rearranging gives
\begin{eqnarray*}
S_{2}(x)=\frac{x^3(1+2x)}{(1-x)^5}. \\
\end{eqnarray*}
Answer $2$: Combinatorial interpretation.
The stirling number ${n \brace k}$ of the second kind are the number of ways to split an $n$-set into $k$ (disjoint) blocks.
For ${n \brace n-2}$ there are two possibilities
$1)$ A block of size $3$ and $n-3$ blocks of size $1$ ... & there are $\binom{n}{3}$ ways to choose these configurations, giving the generating function
\begin{eqnarray*}
\frac{x^3}{(1-x)^4}. \\
\end{eqnarray*}
$2)$ $2$ blocks of size $2$ and $n-4$ blocks of size $1$ ... & there are $3 \times \binom{n}{4}$ ways to choose these configurations, giving the generating function
\begin{eqnarray*}
\frac{3x^4}{(1-x)^5}. \\
\end{eqnarray*}
Now add these two terms to obtain the required generating function.
| {
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"question_score": "2",
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Need Help with Mistake in Generating Function for Nonlinear Recurrence relation I'm having a bit of trouble finding the generating function for the following recurrence relation:
$$
w_n -1 = \sum _{k=1}^{n-1} w_k w_{n-k}, \quad n \geq 2, \; w_0 = 0, \; w_1 = 1.
$$
I set out to find a generating function $F$ such that
$$
F(x) = \sum _{n=0}^{\infty} w_n x^n.
$$
First, I multiplied the LHS of the equation with $x^n$ and summed, obtaining
$$
F(x) - (w_0 + w_1 x) - \sum _{n=0}^{\infty} x^n = F(x)-x-\frac{1}{1-x}.
$$
For the RHS, we can observe that the $n$-th coefficient of $F(x)^2$ concide with
$$
\sum _{k=1}^{n-1} w_k w_{n-k}.
$$
That is, if we perform the product term by term and collect, we can check
$$
F(x)^2 = w_1 w_1 x^2 + (w_1 w_2 + w_2 w_1)x^3 + (w_1 w_3 + w_2 w_2 + w_3 w_1)x^4 + \cdots
$$
and this is possible because $w_0 = 0$. Equating both expressions, I get
$$
F(x) - x - \frac{1}{1-x} = F(x)^2.
$$
But this equation is wrong! Because if I take $x=0$,
$$
(0) - (0) - \frac{1}{1 - (0)} = 0 \implies -1 = 0.
$$
I haven't been able to spot my mistake. Any and all help is appreciated :)
| Multiplying both sides of the recurrence by $x^n$ and summing for $n=2$ to infinity:
$$
\sum_{n=2}^\infty w_nx^n = \sum_{n=2}^\infty \sum_{k=1}^{n-1} w_kw_{n-k}x^n + \sum_{n=2}^\infty x^n.
$$
Because $w_0=0$, $w_1=1$, and $w_n$ is a sum of products of $w_1,\ldots,w_{n-1}$ plus one, $w_n$ is nonnegative for all $n$. So by Tonelli's theorem we may interchange the order of summation:
$$
F(x) - x = \sum_{k=1}^\infty w_k\sum_{n=k+1}^\infty w_{n-k}x^n + \frac{x^2}{1-x}.
$$
Shifting the index of the sum over $k$ down by $n$, we have
$$
F(x) = \sum_{k=1}^\infty w_k x^k\sum_{n=1}^\infty w_nx^n + x + \frac{x^2}{1-x}.
$$
But since $w_0=0$, $\sum_{n=0}^\infty w_nx^n = \sum_{k=1}^\infty w_kx^k = F(x)$, so we have
$$
F(x) = F(x)^2 + x + \frac{x^2}{1-x},
$$
and hence
$$
F(x) - F(x)^2 = x + \frac{x^2}{1-x}.
$$
The roots of this equation are
$$
F(x) = \frac{1}{2} \left(1-\frac{\sqrt{5 x-1}}{\sqrt{x-1}}\right),\quad F(x) = \frac{1}{2} \left(\frac{\sqrt{5 x-1}}{\sqrt{x-1}}+1\right).
$$
Since $F(0) = w_0 = 0$, we see that the first root is the correct expression for $F(x)$, and so
$$
F(x) = \frac{1}{2} \left(1-\frac{\sqrt{5 x-1}}{\sqrt{x-1}}\right).
$$
Unfortunately there is unlikely to be a closed form series expression for $F$. Mathematica only returns something in the form of $\texttt{DifferenceRoot}$ and the first few terms
$$
0,1,2,5,15,51,188,731,2950,12235,51822,223191
$$
didn't match anything on OEIS.
| {
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Determine the conditions for $n$ and $\theta\neq0+2k\pi$ such that $(\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta})^n \in\mathbb{R}$ Let $z=1+\cos\theta+i\sin\theta=|z|(\cos\alpha+i\sin\alpha)$ and $z'=1-\cos\theta+i\sin\theta=|z'|(\cos\alpha'+i\sin\alpha')$.
My line of reasoning is to convert the numerator and denominator to polar form, then knowing that in order for $(\frac{z}{z'})^n$ to be real, we must have $\arg(\frac{z}{z'})^n \equiv 0 \pmod{\pi}$ or more succinctly $$n(\arg(z)-\arg(z')) \equiv 0 \pmod{\pi}$$.
Determining $\arg(z) = \alpha$ and $\arg(z') = \alpha'$: $$\tan{\alpha}=\frac{\sin{\theta}}{1+\cos{\theta}}=\frac{2\sin{\frac{\theta}{2}\cos{\frac{\theta}{2}}}}{2\cos^2{\frac{\theta}{2}}}=\frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}=\tan{\frac{\theta}{2}}$$ thus $\alpha = \frac{\theta}{2}$. $$\tan{\alpha'}=\frac{\sin{\theta}}{1-\cos{\theta}}=\frac{2\sin{\frac{\theta}{2}\cos{\frac{\theta}{2}}}}{2\sin^2{\frac{\theta}{2}}}=\frac{\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=\frac{\sin{\alpha'}}{\cos{\alpha'}}$$ thus $$\sin{\alpha'}\sin{\frac{\theta}{2}}=\cos{\alpha'}\cos{\frac{\theta}{2}} \Leftrightarrow \frac{1}{2}(\cos(\alpha'-\frac{\theta}{2})-\cos(\alpha'+\frac{\theta}{2}))=\frac{1}{2}(\cos(\alpha'-\frac{\theta}{2})+\cos(\alpha'+\frac{\theta}{2}))$$ therefore $$-\cos(\alpha'+\frac{\theta}{2})=\cos(\alpha'+\frac{\theta}{2})$$. Knowing that $$\cos(\pi+\alpha'+\frac{\theta}{2})=\cos(\alpha'+\frac{\theta}{2})$$ is impossible, we take $$\cos(\pi-\alpha'-\frac{\theta}{2})=\cos(\alpha'+\frac{\theta}{2})$$ hence $$\pi-\alpha'-\frac{\theta}{2} = \alpha'+\frac{\theta}{2} \Leftrightarrow \alpha' = \frac{\pi-\theta}{2}$$.
So now we have $$\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta} = \frac{|z|(\cos{\frac{\theta}{2}}+i\sin{\frac{\theta}{2}})}{|z'|(\cos{\frac{\pi-\theta}{2}}+i\sin{\frac{\pi-\theta}{2}})}$$
Here's where I have some difficulties. Supposing
$$\frac{\theta}{2}-\frac{\pi-\theta}{2} \equiv 0 \pmod{\pi}$$ therefore $$\frac{2\theta - \pi}{2} \equiv 0 \pmod{\pi} \Leftrightarrow 2\theta - \pi \equiv 0 \pmod{2\pi}$$ thus $$\theta \equiv \frac{\pi}{2} \pmod{\pi}$$.
However, for all $n$ and using the above condition, $$(\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta})^n = 1$$
It can't be for all $n$, there's something here I'm missing.
| Easier is to simplify the denominator using conjugates. \begin{align*}
&\frac{1 + \cos \theta + \mathrm{i} \sin \theta}{1 - \cos \theta + \mathrm{i} \sin \theta} \cdot \frac{1 - \cos \theta - \mathrm{i} \sin \theta}{1 - \cos \theta - \mathrm{i} \sin \theta} \\
&\quad{}= \frac{2 \sin \theta}{2 - 2 \cos \theta} (-\mathrm{i} \cos \theta + \sin \theta) \text{,}
\end{align*}
which is the product of a real number and a complex number of magnitude $1$. The denominator of the real number explains the constraint in the problem statement, so that division by zero is avoided.
It is worth noting
$$ (-\mathrm{i} \cos \theta + \sin \theta)^n = \frac{(\cos \theta + \mathrm{i} \sin \theta)^n}{\mathrm{i}^n} \text{.} $$
When $n$ is even, this denominator is real, so you only require the numerator be real. When $n$ is odd, the denominator is pure imaginary, so you only require the numerator be pure imaginary...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this a correct proof of the divergence of a series? I wanted to show if
$$ \sum_{n=2}^\infty \frac{\sqrt{n}}{3^n\ln(n)} $$
converges or diverges. By the ratio test, I could solve for $a_n=\frac{\sqrt{n}}{3^n\ln(n)}$ the limit
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}$$
and see what the result tells me.
$Proof:$
$i)$ Let $f(x)=\frac{\sqrt{x}}{3^x\ln(x)}$. Then, because $f(n)=a_n$ and $f(n+1)=a_{n+1}$,
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=\lim_{x\to\infty} \frac{f(x+1)}{f(x)}$$
$ii$) $$\lim_{x\to\infty} \frac{f(x+1)}{f(x)}= \lim_{x\to\infty} \frac{\sqrt{x+1}}{3^{x+1}\ln(x+1)}*\frac{3^x \ln(x)}{\sqrt{x}}=\frac{1}{3}\lim_{x\to\infty} \frac{\sqrt{x+1}}{\sqrt{x}} *\frac{\ln(x)}{\ln(x+1)} $$
$$= \frac{1}{3}\lim_{x\to\infty} \sqrt{\frac{x+1}{x}}*\frac{x+1}{x}$$
$iii)$ It is not hard to show that $\lim_{x\to\infty} \sqrt{\frac{x+1}{x}}$ and $\lim_{x\to\infty} \frac{x+1}{x}$ both exist and are equal to one. Then
$$= \frac{1}{3}\lim_{x\to\infty} \sqrt{\frac{x+1}{x}}*\frac{x+1}{x} = \frac{1}{3}(\lim_{x\to\infty} \sqrt{\frac{x+1}{x}}*\lim_{x\to\infty} \frac{x+1}{x})=\frac{1}{3}$$
Because of $i$ we conclude that $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{1}{3}<1$$
$iiii$) Because the ratio is less than one, then the series converges.
| It's correct. It can also be done without switching to functional limits and using L'Hôpital's rule:
$$\frac{a_{n+1}}{a_n}=\frac 13\sqrt{1+\frac 1n}\frac{\ln n}{\ln(n+1)}\to \frac 13\cdot 1\cdot 1=\frac 13<1 $$
To handle the fraction with the logarithms
$$\frac{\ln(n+1)}{\ln n}=\frac{\ln n+\ln\left(1+\frac 1n\right)}{\ln n}=1+\ln\left(1+\frac 1n\right)\cdot\frac{1}{\ln n}\to1+0\cdot 0=1 $$
so the reciprocal $\ln(n)/\ln(n+1)$ also has limit $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Logarithmic inequality $5\times \frac{3^{x-2}}{3^{x}-2^{x}}\geq1+(\frac{2}{3})^x$ Can somebody help me with this problem?
Solve inequation
$$5\times \frac{3^{x-2}}{3^{x}-2^{x}}\geq1+(\frac{2}{3})^x$$
I am trying to multiply both sides of inequation with
$$\frac{3^{x}}{3^{x}+2^{x}}$$
and I get
$$\frac{5\times3^{2x}}{9\times (3^{2x}-2^{2x})} \geq1$$
but I don t know how to continiue
| You can rewrite it as follows:
$$5 \geq
(1 + \left(\frac{2}{3}\right)^x) \frac{3^x - 2^x}{3^{x-2}}=
(1 + \left(\frac{2}{3}\right)^x) ( 9 - \frac{1}{9} \left(\frac{2}{3}\right)^x) $$
Now if we set $a = \left(\frac{2}{3}\right)^x$, we're reducing the problem to this binomial inequality:
$$(1 + a)(9 - \frac{1}{9}a) \leq 5$$
Knowing the range of $a$, you can obtain the range of $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$ Is there closed form for
$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$$
where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.
My approach,
In this paper page $95$ Eq $(5)$ we have
$$\sum_{n=1}^\infty \overline{H}_n\frac{x^n}{n}=\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2(-x)-\ln2\ln(1-x)-\operatorname{Li}_2\left(\frac12\right)$$
Divide both sides by $x$ then integrate we get
$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n=\int\frac{\operatorname{Li}_2\left(\frac{1-x}{2}\right)}{x}\ dx-\operatorname{Li}_3(-x)+\ln2\operatorname{Li}_2(x)-\operatorname{Li}_2\left(\frac12\right)\ln x$$
and my question is how to find the remaining integral? Thanks
Maybe you wonder why I have it as an indefinite integral, I meant so as I am planning to plug $x=0$ to find the constant after we find the closed form of the integral if possible.
I tried Mathematica, it gave
Edit
With help of $Mathematica$ I was able to find
\begin{align}
\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n&=-\frac13\ln^3(2)+\frac12\ln^2(2)\ln(1-x)-\frac12\zeta(2)\ln(x)+\frac32\ln^2(2)\ln(x)\\
&\quad-\ln(2)\ln(x)\ln(1-x)-\frac12\ln(2)\ln^2(x)-\frac12\ln^2(2)\ln(1-x)\\
&\quad-\ln^2(2)\left(\frac{x}{1+x}\right)+\ln(2)\ln\left(\frac{x}{1+x}\right)[\ln(1-x)+\ln(x)]\\
&\quad+\ln(x)\ln(1-x)\ln(1+x)+\ln(x)\operatorname{Li}_2\left(\frac{1-x}{2}\right)+\ln\left(\frac{x}{1+x}\right)\operatorname{Li}_2(x)\\
&\quad+\ln(1+x)\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{x}{1+x}\right)\ln\left(\frac{2x}{1+x}\right)-\operatorname{Li}_2\left(\frac{2x}{1+x}\right)\ln\left(\frac{2x}{1+x}\right)\\
&\quad+\operatorname{Li}_2\left(\frac{1+x}{2}\right)\ln\left(\frac{x}{2}\right)-\ln\left(\frac{x}{1+x}\right)\operatorname{Li}_2\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(x)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)\\
&\quad+\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(-x)+\ln(2)\operatorname{Li}_2(x)+\frac{7}{8}\zeta(3)
\end{align}
| From this paper page $101$ we have
$$\sum_{n=1}^\infty\overline{H}_n\frac{x^{n+1}}{(n+1)^2}=\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(x)$$
$$+\ln(1+x)\left[\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{1}{2}\right)+\frac12\ln 2\ln(1+x)\right]+\operatorname{Li}_3\left(\frac{1}{2}\right)$$
but
$$\sum_{n=1}^\infty\overline{H}_n\frac{x^{n+1}}{(n+1)^2}=\sum_{n=0}^\infty\overline{H}_n\frac{x^{n+1}}{(n+1)^2}=\sum_{n=1}^\infty\overline{H}_{n-1}\frac{x^n}{n^2},\quad \overline{H}_{n-1}=\overline{H}_n+\frac{(-1)^n}{n}$$
$$=\sum_{n=1}^\infty\overline{H}_{n}\frac{x^n}{n^2}+\operatorname{Li}_3(-x)$$
Thus
$$\sum_{n=1}^\infty\overline{H}_{n}\frac{x^n}{n^2}=\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(-x)-\operatorname{Li}_3(x)$$
$$+\ln(1+x)\left[\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{1}{2}\right)+\frac12\ln 2\ln(1+x)\right]+\operatorname{Li}_3\left(\frac{1}{2}\right)$$
| {
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For △ABC, prove $\frac a{h_a} + \frac b{h_b} + \frac c{h_c} \ge 2 (\tan\frac{\alpha}2+ \tan\frac{\beta}2 + \tan\frac{\gamma}2)$
Given $\triangle ABC$, (using the main parameters and notation), prove that $$ \frac{a}{h_a} + \frac{b}{h_b} + \frac{c}{h_c} \ge 2 \cdot \left(\tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \tan\frac{\gamma}{2}\right)$$
Great, another problem that already has a solution. We have many ways of expressing the area of a triangle, for example $$2 \cdot [ABC] = \frac{abc}{2R} = \sqrt{2(a + b + c) \cdot \sum_{cyc}\frac{c + a - b}{2}} = ah_a = bh_b = ch_c$$
The above equations are used in the solution I have provided below.
I would be greatly appreciated if you could come up with any other solutions.
| Using the double-angle and half-angle trigonometric identities, we have that $$2 \cdot \sum_{cyc}\tan\frac{\beta}{2} = 2 \cdot \sum_{cyc}\frac{\sin\beta}{\cos^2\dfrac{\beta}{2}} = \frac{1}{R} \cdot \sum_{cyc}\frac{b}{\cos\beta + 1} = \frac{2abc}{R} \cdot \sum_{cyc}\frac{1}{(c + a)^2 - b^2}$$
$$ = \frac{8 \cdot [ABC]}{a + b + c} \cdot \sum_{cyc}\frac{1}{c + a - b} = \frac{1}{2 \cdot [ABC]}\cdot \sum_{cyc}(a + b - c)(b + c - a)$$
$$ = \frac{1}{2\cdot [ABC]} \cdot \sum_{cyc}[b^2 - (c - a)^2] \le \frac{a^2 + b^2 + c^2}{2 \cdot [ABC]} = \sum_{cyc}\frac{b}{h_b}$$
| {
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Evaluating $P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ when $a+b+c=0$
Let $a,b,c$ such that $$a + b + c =0$$ and
$$P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$$ is defined.
Find the value of $P$.
This is a very queer problem.
| Put $c=-a-b$ in
$$2a^2+bc=2a^2+b(-a-b)=a^2-ab+a^2-b^2=(a-b)(2a+b)=(a-b)(b-c)$$
Similarly $2b^2+ca=(b-c)(c-a)$ and
$2c^2+ab=(c-a)(c-b)$
So expression reduce into
$$\frac{a^2(c-a)+b^2(a-b)+c^2(b-c)}{(a-b)(b-c)(c-a)}=1$$
Factorisation of Numerator
$a^2(c-a)+b^2(a-b)+c^2(b-c)$ is zero when $a=b$ or $b=c$ or $c=a$
$$a^2(b-c)+b^2(c-a)+c^2(a-b)=k(a-b)(b-c)(c-a)$$
For finding $k$ put $a=1,b=2,c=3,$ getting $k=1$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I calculate this (rather tricky) limit? $$\lim\limits_{n \to \infty} \cos^{n^2} \left (\frac{2x}{n} \right)$$
Any hints and/or help is greatly appreciated.
| Consider $$a_n=\cos^{n^2} \left (\frac{2x}{n} \right)\implies \log(a_n)=n^2\log\left( \cos\left (\frac{2x}{n} \right)\right)$$ Now use the series expansion
$$\cos\left (\frac{2x}{n} \right)=1-\frac{2 x^2}{n^2}+\frac{2 x^4}{3 n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log\left( \cos\left (\frac{2x}{n} \right)\right)=-\frac{2 x^2}{n^2}-\frac{4 x^4}{3 n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log(a_n)=-2 x^2-\frac{4 x^4}{3 n^2}+O\left(\frac{1}{n^4}\right)$$
$$a_n=e^{\log(a_n)}=e^{-2 x^2}\left( 1-\frac{4 x^4}{3 n^2}\right)+O\left(\frac{1}{n^4}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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integration of. $\int \frac{x}{x^3-3x+2}$ I am trying to integrate :
$\Large \int \frac{x}{x^3-3x+2}dx$
I decomposed the fraction and got :
$\Large \frac{x}{x^3-3x+2} = \frac {x}{(x-1)^2(x+2)}$
Then I tried to get two different fractions:
$ \Large \frac {x}{(x-1)^2(x+2)} = \frac {Ax}{(x-1)^2} + \frac {B}{(x+2)}$
Well I got A=1/3 and B=-1/3. as possible values. but that was not correct. I guess I missed something.
| When the denominator of a fraction has an irreducible factor $\bigl(p(x)\bigr)^m$ with order of multiplicity $m>1$, its contribution to the decomposition into partial fractions is not
$$\frac{A(x)}{\bigl(p(x)\bigr)^m} \qquad (\deg A(x)<\deg p(x)),$$
but
$$\frac{A_1(x)}{p(x)\vphantom{\Big)}}+\frac{A_2(x)}{\bigl(p(x)\bigr)^2}+\dots+\frac{A_m(x)}{\bigl(p(x)\bigr)^m}\\(\deg A_1(x),\deg A_2(x),\dots,\deg A_m(x)<\deg p(x)) $$
Therefore here, as the irreducible factors have degree $1$, the decomposition has the form
$$\frac{x}{x^3-3x+2} = \frac {x}{(x-1)^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}.$$
Addendum:
Here is the short way to determine the coefficients: multiply both sides of the above equality by $(x-1)^2(x+2)$ to obtain the relation
$$x=(x-1)(x+2)A+(x+2)B+(x-1)^2C.$$
Setting $x=1$, then $x=-2$, yields instantly
$B=\frac 13$ and $C=-\frac29$. To obtain $C$, observe that the r.h.s. must have no quadratic term, so $A+C=0$ and finally $A=\frac29$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimize $\frac{2}{1-a}+\frac{75}{10-b}$
Let $a,b>0$ and satisfy $a^2+\dfrac{b^2}{45}=1$. Find the minimum
value of $\dfrac{2}{1-a}+\dfrac{75}{10-b}.$
WA gives the result that $\min\left(\dfrac{2}{1-a}+\dfrac{75}{10-b}\right)=21$ with $a=\dfrac{2}{3},b=5$.
Consider making a transformation like this
$$\dfrac{2}{1-a}+\dfrac{75}{10-b}=\frac{2\cdot (ma)^2}{ma\cdot ma\cdot (1-a)}+\frac{75\cdot(nb)^2}{nb\cdot nb\cdot(10-b)}.$$
If $2m=2n=1$, $ma=1-a$, and $ nb=10-b$ can all hold, we can apply AM-GM inequality as follows
$$\frac{2\cdot (ma)^2}{ma\cdot ma\cdot (1-a)}+\frac{75\cdot(nb)^2}{nb\cdot nb\cdot(10-b)}\geq \frac{2\cdot (ma)^2}{\left(\frac{ma
+ma+(1-a)}{3}\right)^3}+\frac{75\cdot(nb)^2}{\left(\frac{nb+nb+(10-b)}{3}\right)^3}=\cdots$$
But this is invalid, since $n=\frac{1}{2},nb=10-b$ can not satisfy $b=5$.
How to solve it?
| There is another way using parameterisation. Let $a = \cos t, b = \sqrt{45} \sin t$, then we have:
$$f(t) = \frac{2}{1-\cos t} + \frac{75}{10-\sqrt{45} \sin t}$$
$$f'(t) = -\frac{2 \sin t}{(1-\cos t)^2} + \frac{225 \sqrt5 \cos t}{(10-3 \sqrt{5} \sin t)^2}$$
The condition that $a,b > 0$ translates to $0 < t < \frac{\pi}{2}$. Setting $f'(t)$ equal to $0$, we find that only $0.841$ (actually $\tan^{-1} \left( \frac{1}{\sqrt5} \right)$) is inside the range.
Therefore the minimum value is $f(0.841) = 21$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the remainder when $x^{100}$ is divided by $x^8 - x^6 + x^4 - x^2 + 1.$ I need help in the problem:
Find the remainder when $x^{100}$ is divided by $x^8 - x^6 + x^4 - x^2 + 1.$
I have tried factoring $x^{100}-1$ and adding 1 to that, but that hasn't helped. Could someone please help with this?
| Hint
Replace $x^2$ with to find the divisor $$=\dfrac{y^5+1}{y+1}$$
$y^{50}=(y^5+1-1)^{10}=1+$ terms divisible by $y^5+1$
| {
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Proof by induction:$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$ In the very beginning I'm going to refer to similar posts with provided answers:
Induction Inequality Proof with Product Operator $\prod_{k=1}^{n} \frac{(2k-1)}{2k} \leq \frac{1}{\sqrt{3k+1}}$ (answered by Özgür Cem Birler)
Prove that $\prod\limits_{i=1}^n \frac{2i-1}{2i} \leq \frac{1}{\sqrt{3n+1}}$ for all $n \in \Bbb Z_+$
I examined the solutions and tried to apply the methods used there to make sure whether I understand it or not. I'm concerned about the step of induction.
A task from an earlier exam at my university:
Prove by induction:
$$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$$
Attempt:
rewritten:
$$\prod_{i=1}^n\frac{4i-1}{4i+1}<\sqrt{\frac{3}{4n+3}}$$
$(1)$ base case: $\tau(1)$
$$\frac{3}{5}=\sqrt{\frac{4}{7}}<\sqrt{\frac{3}{7}}$$
$(2)$ assumption:
Let:$$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$$
hold for some $n\in\mathbb N$
$(3)$ step: $\tau(n+1)$
$$\frac{4n+3}{4n+5}\cdot\prod_{i=1}^n\frac{4i-1}{4i+1}<\frac{4n+3}{4n+5}\cdot\sqrt{\frac{3}{4n+3}}=\frac{\sqrt{3(4n+3)}}{4n+5}<\sqrt{\frac{3}{4n+7}}$$
$$\frac{12n+9}{16n^2+40n+25}<\frac{3}{4n+7}\iff\frac{48n^2+120n+63-48n^2-120n-75}{\underbrace{(4n+5)^2(4n+7)}_{>0}}<0\iff-\frac{12}{(4n+5)^2(4n+7)}<0$$
Is this combined legitimate?
| I concur with everyone else that it's basically right. Everyone has their own style, but the following is how I would probably write up the "meat and potatoes" of the proof:
Let's start with some preliminary observations. Note that
\begin{align}
\frac{4n+3}{4n+5}\cdot\sqrt{\frac{3}{4n+3}}
&=\frac{4n+3}{4n+5}\cdot\frac{\sqrt{3}}
{\sqrt{4n+3}}\\[1em]
&=\frac{\sqrt{4n+3}\cdot\sqrt{3}}{4n+5}\\[1em]
&= \sqrt{\frac{(4n+3)(3)}{(4n+5)^2}}\\[1em]
&= \sqrt{\frac{12n+9}{16n^2+40n+25}}.
\end{align}
More concisely, we have
$$
\frac{4n+3}{4n+5}\cdot\sqrt{\frac{3}{4n+3}}=\sqrt{\frac{12n+9}{16n^2+40n+25}}.\tag{1}
$$
As another observation, note that
\begin{align}
\frac{12x+9}{16x^2+40x+25} < \frac{3}{4x+7}
&\iff \frac{12x+9}{16x^2+40x+25} - \frac{3}{4x+7} < 0\\[1em]
&\iff \frac{(48x^2+120x+63)-(48x^2+120x+75)}{(4x+5)^2)(4x+7)}<0\\[1em]
&\iff \frac{-12}{(4x+5)^2(4x+7)}<0\\[1em]
&\iff \frac{12}{(4x+5)^2(4x+7)}>0\\[1em]
&\iff x\in\Bigl(-\frac{7}{4},-\frac{5}{4}\Bigr)\cup\Bigl(-\frac{5}{4},\infty\Bigr).
\end{align}
More specifically, note that, for any natural number $n$, it follows from above that
$$
\frac{12n+9}{16n^2+40n+25} < \frac{3}{4n+7}.\tag{2}
$$
Now your proof can flow very naturally:
\begin{align}
\prod_{i=1}^{k+1}\frac{4i-1}{4i+1}
&=\prod_{i=1}^k\frac{4i-1}{4i+1}\cdot\frac{4(k+1)-1}{4(k+1)+1}\\[1em]
&<\frac{4k+3}{4k+5}\cdot\sqrt{\frac{3}{4k+3}}\\[1em]
&=\sqrt{\frac{12k+9}{16k^2+40k+25}} & \text{(by $(1)$)}\\[1em]
&<\sqrt{\frac{3}{4k+7}} & (\text{$\sqrt{x}$ strictly increases and by $(2)$})\\[1em]
&= \sqrt{\frac{3}{4(k+1)+3}}. & \text{(desired conclusion)}
\end{align}
Maybe that's overdrawn and slightly verbose, but that's how I would write it up if I were doing it for a class.
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"answer_id": 1
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Show that $a^2$ cannot be congruent to 2 or 3 mod 5 for any integer a) Show that $a^2$ cannot be congruent to $2$ or $3 \bmod 5$ for any integer.
$a^2≡2,3 \pmod5$
$a≡0,±1,±2\pmod5 )⟹a^2≡0,1,4$.
But $2,3$ are not congruent to $0,1,4\pmod5$.
I am not sure if I did it right, please check it for me
(b) Show that if $5\mid x^2+y^2+z^2$ then either exactly one of, or each of $x, y, z$ are a multiple of $5$.
I really can not figure it out :(
| You did a) just find. As $(-2,-1,0,1,2)$ is a complete residue class $\mod 5$ it must be that $a\equiv 0,\pm 1, \pm 2\pmod 5$ and therefore $a^2\equiv 0, 1,4\equiv -1$ none are congruent to $2$ or to $3\equiv -1$.
As for $b$ we have $x^2,y^2, z^2 \equiv 0, 1,-1\mod 5$ so we mus have some how that $x^2 + y^2 + z^2 \equiv (0|1|-1) + (0|1|-1) + (0|1|-1)\equiv 0$.
And it should be clear of all the combintations:
$0 + 0 + 0 \equiv 0$ (Hit)
$0 + 0 + 1\equiv 1$,$0+0-1\equiv -1; 0+1+1\equiv 2$ (Misses)
$0 + 1\equiv -1\equiv 0$ (Hit)
$0 -1 -1\equiv -2; 1+1+1\equiv 3; 1+1-1\equiv 1; 1-1-1\equiv -1; -1-1-1\equiv -3$ (Misses)
The only way $x^2 + y^2 + z^3 \equiv 0\pmod 5$ is if all $x^2,y^2, z^2 \equiv 0$ in which cae $x,y, z\equiv 0$ and are all multiples of $5$
or if one of them is a multiple of $5$ and the other two .... are not. (Note, we never said that the other two could be anything; we only said they can't be multiples of $5$. One must be $x$ (or $y$) $ \equiv \pm 1$ so $x^2$( or $y^2$) $\equiv 1$ and the other must be $y$ (or $x$) $\equiv \pm 2$ so that $y^2$ (or $x^2$) $ \equiv 4\equiv -1$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove the inequality stated below:
If $a, b, c, d > 0$ then prove:
$$\frac{a}{b+c+d} + \frac{b+c}{a+d} + \frac{d}{b+a} > 1$$
I tried to consider the difference and lead to a common denominator, but in the numerator, there remains ${1}$ term with a minus which is less than ${0}$
$a^3 - a^2 c + b (b + c)^2 + c d^2 + d^3 + a (b + c) (c + d)$
| Notice that if $m,n,k>0$ then $${m\over n} >{m\over n+k}$$ so
$$\frac{a}{b+c+d} + \frac{b+c}{a+d} + \frac{d}{b+a} >\frac{a}{\color{red}{a}+b+c+d} + \frac{b+c}{\color{red}{b+c}+a+d} + \frac{d}{\color{red}{c+d}+b+a}$$ $$= \frac{a+b+c+d}{a+b+c+d}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3540178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$
What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$
We have to make this function defined, by change the cosine into sine or tan. So,
$$\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{2\sin 3t \cos 3t}{t^3 \cos^2 3t}\right) = \lim_{t \to 0} \left(\frac{a}{t^2} - \frac{6}{t^2}\right) = -18$$
But I am confused what should I do next.
| Just use Taylor expansion and simplify first $\cos^2 3t = \frac{1+\cos 6t}{2}$.
Hence,
$$\frac{at\cos^2 3t - \sin 6t}{t^3\cos^2 3t}= \frac{at(1+\cos 6t) - 2\sin 6t}{t^3(1+\cos 6t)}$$
$$= \frac{at + at(1- 18t^2+o(t^3)) - 2(6t-36t^3+o(t^4))}{t^3(1+\cos 6t)}$$
$$= \frac{1}{(1+\cos 6t)}\left(\frac{2a-12}{t^2} -18a+72 + o(t)\right)$$
So, the limit exists only if $\boxed{a=6}$ and then the limit is equal to $\boxed{\frac{-36}{2}}=-18$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Solve $xy''-y'-x^3y=0$ I want to solve $xy''-y'-x^3y=0$. My solution:
$y = z\cdot \exp(kx^2)$
$y' = z'\cdot \exp(kx^2) + z\cdot 2kx\exp(kx^2)$
$y'' = z''\cdot \exp(kx^2) + z'\cdot 4kx\exp(kx^2) + z\cdot (2k+4k^2x^2)\exp(kx^2)$
Plug in:
$z''\exp(kx^2)+z'(4kx-\frac{1}{x})\exp(kx^2)+z(2k+4k^2x^2-\frac{1}{x}\cdot2kx-x^2)\exp(kx^2)=0
$
$2k+4k^2x^2-\frac{1}{x}\cdot2kx-x^2=0\iff k=\frac{1}{2}$
Plug in:
$z''+z'(2x-x^{-1})=0$
$z'=:u$
$u'+u(2x-x^{-1})=0$
$\frac{\mathrm{d}u}{u}=-(2x-x^{-1})\mathrm{d}x$
$\ln|u|=-(x^2-\ln|x|)+c_1$
$u=\frac{c_1x}{e^{x^2}}$
Substitute back:
$z'=c_1xe^{-x^2}$
$z=-\tfrac{1}{2}c_1e^{-x^2}+c_2$
$y(x)=(-\tfrac{1}{2}c_1e^{-x^2}+c_2)e^{\frac{1}{2}x^2}$
$y(x)=c_2e^{\frac{1}{2}x^2}-\tfrac{1}{2}c_1e^{-\frac{1}{2}x^2}$
Right? Is there another way solving this ode? Thanks!
edit: @Lutz Lehmann: Sorry there was a square missing. Fixed it.
| Dividing by $x^2$, the equation is
$$\left(\frac{y'}x\right)'-xy=0$$
or
$$\frac{y'}x\left(\frac{y'}x\right)'-yy'=0.$$
After integration,
$$\left(\frac{y'}x\right)^2-y^2=c,$$
which is separable.
$$\frac{y'}{\sqrt{y^2+c}}=\pm x$$
gives
$$\text{arsinh}\frac y{\sqrt{|c|}}=c'\pm\frac{x^2}2$$ or
$$\text{arcosh}\frac y{\sqrt{|c|}}=c'\pm\frac{x^2}2$$ depending on the sign of $c$.
Final form:
We can rewrite as
$$y=\frac{\sqrt{|c|}}2\left(e^{c'+x^2/2}\pm e^{-c'-x^2/2}\right)$$
or
$$y=c_+e^{x^2/2}+c_-e^{-x^2/2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation $\sqrt[3]{15-x^3+3x^2-3x}=2\sqrt{x^2-4x+2}+3-x$. Solve the equation $\sqrt[3]{15-x^3+3x^2-3x}=2\sqrt{x^2-4x+2}+3-x$.
I have tried to solve for x by Casio and try to make the equation to $u.v=0$ but the solution is not in $\mathbb{Q}$. Any help is appreciated. Thanks
| For the square root to be defined we need:
$$x^2-4x+2\geq 0$$
Therefore, we have:
$$2\sqrt{x^2-4x+2} = x-3+\sqrt[3]{15-x^3+3x^2-3x}=$$
$$=\frac{(x-3)^3+15-x^3+2x^2-3x}{(x-3)^2-(x-3)\sqrt[3]{15-x^3+3x^2-3x}+\sqrt[3]{(15-x^3+3x^2-3x)^2}}$$
$$=\frac{-6(x^2-4x+2)}{(x-3)^2-(x-3)\sqrt[3]{15-x^3+3x^2-3x}+\sqrt[3]{(15-x^3+3x^2-3x)^2}}\leq 0$$
The numerator is negative $x^2-4x+2 \geq 0 \Rightarrow -6(x^2-4x+2)\leq 0$. The denominator is of the form $a^2-ab+b^2$ which is always non-negative because:
$$a^2-ab+b^2=\frac{1}{2}[a^2+b^2+(a-b)^2]\geq 0$$
And thus $x^2-4x+2=0$ which means $x=2\pm\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3545501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$$
$$h(x)=\dfrac{x^2+x^2-2+2}{x-\dfrac{1}{x}}$$
$$h(x)=x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}$$
Case $1$: $x-\dfrac{1}{x}>0$
$$\dfrac{x^2-1}{x}>0$$
$$x\in(-1,0) \cup (1,\infty)$$
$$AM\ge GM$$
$$\dfrac{x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}}{2}>\sqrt{2}$$
$$h(x)\ge 2\sqrt{2}$$
$$x-\dfrac{1}{x}=\dfrac{2}{x-\dfrac{1}{x}}$$
$$x^2+\dfrac{1}{x^2}-2=2$$
$$x^2+\dfrac{1}{x^2}=4$$
$$x^4-4x^2+1=0$$
$$x^2=2\pm\sqrt{3}$$
Only $x=\sqrt{2+\sqrt{3}},-\sqrt{2-\sqrt{3}}$ are the valid solutions.
Case $2$: $x-\dfrac{1}{x}<0$
$$x\in(-\infty,-1) \cup (0,1)$$
$$h(x)=-\left(\dfrac{1}{x}-x+\dfrac{2}{\dfrac{1}{x}-x}\right)$$
By $AM\ge GM$, $h(x)\ge-2\sqrt{2}$
We will get this minimum value at $-\sqrt{2+\sqrt{3}},\sqrt{2-\sqrt{3}}$
So answer should have been $-2\sqrt{2}$ but actual answer is $2\sqrt{2}$. What am I missing here.
| Assuming that h(x)= f(x)/g(x), I would not actually calculate h itself. Instead, use the "quotient rule": $\frac{h'(x)= f'(x)g(x)- f(x)g'(x)}{g^2(x)}$. That will be 0 if and only if the numerator is 0: $f'(x)g(x)- f(x)g'(x)= 0$.
$f(x)= x^2+ x^{-2}$ so $f'(x)= 2x+ 2x^{-3}$, $g(x)= x- x^{-1}$ so $g'(x)= 1- x^{-2}$.
So $f'(x)g(x)- f(x)g'(x)= (2x+ 2x^{-3})(x- x^{-1})- (x^2+ x^{-2})(1- x^{-2})= 0$
Seeing that "$x^{-3}$ and $x^{-1}$ I would now multiply both sides by $x^4$ (of course x cannot be 0):
$(2x^4+ 2)(x^2- 2)- (x^4+ 1)(x^2- 1)= 2x^6- 4x^4+ 2x^2- 4- x^6+ x^4- x^2- 1= -3x^4+ x^3- 5= 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
The area of interior quadrilateral formed by connecting trisect points and vertices of larger quadrilateral I drew a figure on GeoGebra to explore the area of the smaller quadrilateral formed by joining the vertices of the larger quadrilateral and the trisect points on the edges of the larger quadrilateral.
Here is what I think to be correct:
$$Area_{\Delta E_2H_2G_2F_2}=\frac{2}{5}Area_{\Delta A_1B_1C_1D_1}$$
I haven't figured out a way to prove this question yet. If this is correct, could any of you provide a possible solution to the problem?
| Some example caculation for retangular case. Let $ABCD$ be a rectangular with $AB=6,AD=3$.
Let $TQ=x$,$XM=y$ and $\angle ZTX=\alpha$, then $XM=ZP=x$; $TQ=YN=y$; $DX=BZ=3x$;$AT=CY=3y$ and $XYZT$ is parallelogram rhombus with $XY=6x$ and $XT=6y$
We have $37a^2=(6a)^2+a^2=AB^2+AQ^2=BQ^2=100x^2$, then $x=\frac{\sqrt{37}}{10}a$
$13a^2=(2a)^2+(3a)^2=AD^2+DM^2=AM^2=(10y)^2=100y^2$, then $y=\frac{\sqrt{13}}{10}a$
and $cos(\alpha)=\frac{TQ^2+TA^2-AQ^2}{2TQ.TA}=\frac{x^2+9y^2-a^2}{6xy}=\frac{9}{\sqrt{481}}$, so $sin(\alpha)=\frac{20}{\sqrt{481}}$
Then $S_{XYZT}=sin(\alpha)XY.XT=\frac{20}{\sqrt{481}}.36\frac{\sqrt{481}a^2}{100}=7.2a^2$, and $S_{ABCD}=18a^2$ so $\frac{S_{XYZT}}{S_{ABCD}}=\frac{2}{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Trouble with trig substitution I have a few questions and a request for an explanation.
I worked this problem for a quite a while last night. I posted it here.
Here is the original question: $$\int\frac{-7 x^2}{\sqrt{4x-x^2}} dx$$
And here is the work that I did on it:
Help with trig sub integral
Sorry that the negative sometimes gets cut off in the photo, and yes I know it's not fully simplified there.
\begin{align}
-7 \int\frac{x^2}{\sqrt{4x-x^2}} dx
&= -7 \int\frac{x^{3/2}}{\sqrt{4-x}} dx \\
&= -7 \int\frac{8\sin^3\theta\ 2\cos\theta}
{\sqrt{4-4\sin^2\theta}} d\theta
&& \begin{array}{c}
2\sin\theta = \sqrt x \\
2\cos\theta = dx \\
(2\sin\theta)^3 = x^{3/2}
\end{array} \\
&= -7 \cdot 8 \int\frac{\sin^3\theta\ 2\cos\theta}
{2\sqrt{1 - \sin^2\theta}} d\theta \\
&= -56 \int \sin^3\theta\, d\theta \\
&= -56 \int (1 - \cos^2\theta) \sin\theta\, d\theta \\
&= -56 \int(\sin\theta - \cos^2\theta\sin\theta)\,d\theta \\
&= 56 \cos\theta - 56 \int u^2\, du &&
\begin{array}{c}
u = \cos\theta \\
du = \sin\theta
\end{array}\\
&= 56 \cos\theta - 56 \frac{\cos^3\theta}{3} + C \\
&= 56 \left(\frac{\sqrt{4-x}}{2}
- \frac13 \left( \frac{\sqrt{4-x}}{2}\right)^3 \right)+ C
&& \cos\theta = \frac{\sqrt{4-x}}{2}\\
&= 56 \frac{\sqrt{4-x}}{2}
\left(1 - \frac{4-x}{12}\right) + C\\
\end{align}
My first question is the more involved one: Is the algebra in my original work sound? If it is, why doesn't it work in this instance?
My second question is: is this a correct solution?
$$14\left(\frac{\sqrt{4x-x^2}(x-2)}{2}-2\sqrt{4x-x^2}-3\arcsin\left(\frac{x-2}{2}\right)\right)+C$$
It is for webwork, and I used two out of three chances. I'd prefer to keep my perfect webwork mark, obviously ;p
Finally, I was kind of impressed with Ans4's square completion and had to run it through to see that it was correct. That's such a useful skill. Do you have some specific advice about how I could improve my math tricks to that point?
| The mistake in your original attempt has been identified:
the substitution for $dx.$
In particular,
$$ \frac{d}{d\theta} 4 \sin^2 \theta = 8 \sin\theta\cos\theta \neq 2 \cos\theta.$$
Patching your original attempt by making the correct substitution,
\begin{align}
-7 \int\frac{x^2}{\sqrt{4x-x^2}} dx
&= -7 \int\frac{x^{3/2}}{\sqrt{4-x}} dx \\
&= -7 \int\frac{8\sin^3\theta\ 8 \sin\theta\cos\theta}
{\sqrt{4-4\sin^2\theta}} d\theta
&& \begin{array}{c}
2\sin\theta = \sqrt x \\
8 \sin\theta\cos\theta \,d\theta = dx \\
(2\sin\theta)^3 = x^{3/2}
\end{array} \\
&= -7 \cdot 8 \int\frac{\sin^3\theta\ 8 \sin\theta\cos\theta}
{2\sqrt{1 - \sin^2\theta}} d\theta \\
&= -224 \int \sin^4\theta\, d\theta \\
&= -224 \int (1 - \cos^2\theta) \sin^2\theta\, d\theta \\
&= -224 \int(\sin^2\theta - \cos^2\theta\sin^2\theta)\,d\theta \\
\end{align}
Here we have to switch strategies due to the extra factor $\sin\theta.$
\begin{align}
\int\sin^2\theta \,d\theta &= \int \frac12 (1 - \cos(2\theta)) \,d\theta\\
&= \frac12\theta - \frac12 \int \cos(2\theta) \,d\theta\\
&= \frac12\theta - \frac14 \sin(2\theta) + C_1
\end{align}
\begin{align}
\int\sin^2\theta\cos^2\theta \,d\theta &= \int \frac14 \sin^2(2\theta) \,d\theta\\
&= \frac18 \int \sin^2 u \,du && u = 2\theta\\
&= \frac18 \left(\frac12u - \frac14 \sin(2u)\right) + C_2 \\
&= \frac18\theta - \frac1{32} \sin(4\theta) + C_2
\end{align}
So we get that
\begin{align}
-224 \int \sin^4\theta\, d\theta
&= -224\left(\frac12\theta - \frac14 \sin(2\theta) + C_1
- \left(\frac18\theta - \frac1{32} \sin(4\theta) + C_2 \right)\right)\\
&= -224\left(\frac38\theta - \frac14 \sin(2\theta)
+ \frac1{32} \sin(4\theta)\right) + C\\
&= -7\left(12\theta - 8\sin(2\theta) + \sin(4\theta)\right) + C\\
&= -7\left(12\theta - 16\sin\theta\cos\theta
+ 2\sin(2\theta)\cos(2\theta)\right) + C\\
&= -7\left(12\theta - 16\sin\theta\cos\theta
+ 4\sin\theta\cos\theta(1-2\sin^2\theta)\right) + C\\
&= -7\left(12\theta - 12\sin\theta\cos\theta
- 8\sin^3\theta\cos\theta\right) + C\\
&= -7\left(12\theta - 12\sin\theta\cos\theta
- 8\sin^3\theta\cos\theta\right) + C\\
&= -7\left(12\arcsin\left(\frac{\sqrt x}2\right) - 6\sqrt x \sqrt{1 - \frac x4}
- x^{3/2}\sqrt{1 - \frac x4}\right) + C\\
&= -7\left(12\arcsin\left(\frac{\sqrt x}2\right) - 3\sqrt{4x - x^2}
- \frac12 x\sqrt{4x - x^2}\right) + C\\
\end{align}
You might recognize that the terms here are similar to the terms in your proposed solution
$$14\left(\frac{\sqrt{4x-x^2}(x-2)}{2}-2\sqrt{4x-x^2}-3\arcsin\left(\frac{x-2}{2}\right)\right)+C,$$
especially if you realize that
$$ \arcsin\left(\frac{\sqrt x}{2}\right)
= \frac12 \arcsin\left(\frac{x-2}{2}\right) + \frac\pi4, $$
but the coefficients do not match.
The proof of the pudding is to take the derivative.
(I did this for both solutions--and when I first did it on my calculations it showed me I had made an arithmetic error, which I was then able to find and correct to get the solution shown here now.)
By the way, I found this a tedious and error-prone approach. Other solutions offer better approaches. I was just curious to see how your original attempt would work out without mistakes.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Increasing each element of a set, what is the sum of the medians of the two sets. Consider an ordered set of six consecutive integers in increasing order. To create a new set of six integers, the first, third, and fifth elements are each multiplied by two, and the second, fourth, and sixth elements are each increased by two. The median of this new set of four more than the median of the first set. What is the sum is the median of the first set and the median of the second set?
If the first number is $a$, then the six numbers are $a, a+1, a+2, a+3, a+4, a+5$. I am thinking that we should have two cases, the first number is odd and the first number is even. However, in those cases, I can't figure out the medians of each of the numbers, so that's where I need help.
| The original sequence of numbers is $a,a+1,a+2,a+3,a+4,a+5$, where $a$ is a positive integer. Let $M_1$ be the median of the old sequence. Note that $M_1=a+2.5$.
The new sequence of numbers is $2a,a+3,2a+4,a+5,2a+8,a+7$. Let $M_2$ be the median of the new sequence.
Note that $2a$ will eventually be bigger than $a+7$, in which case the median will be the mean of $2a$ and $a+7$. Note that
$$2a\ge a+7\quad\text{iff}\quad a\ge7.$$
So if $a\ge7$, then $M_2$ is the mean of $a+7$ and $2a$. So in this case, we'd have that $M_2=1.5a+3.5$. Note that in this case we have that
$$M_2-M_1=0.5a+1\ge4.5.$$
It follows that we must have $a\le6$.
It is now straightforward to check that:
$$\begin{align*}
\text{If }a=1,\text{ then }M_1=3.5&\text{ and }M_2=6.\\
\text{If }a=2,\text{ then }M_1=4.5&\text{ and }M_2=7.5.\\
\text{If }a=3,\text{ then }M_1=5.5&\text{ and }M_2=9.\\
\text{If }a=4,\text{ then }M_1=6.5&\text{ and }M_2=10.\\
\text{If }a=5,\text{ then }M_1=7.5&\text{ and }M_2=11.\\
\text{If }a=6,\text{ then }M_1=8.5&\text{ and }M_2=12.5.
\end{align*}$$
It follows that $a=6$, $M_1=8.5$, $M_2=12.5$, and $M_1+M_2=21$.
| {
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"url": "https://math.stackexchange.com/questions/3548177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does the joint distribution of $X$ and $Y$ is the standard bivariate normal distribution using the change of variables We have $$ X = \sqrt{-2 \log(U)} \cos(2 \pi V)$$ and $$Y = \sqrt{-2 \log(U)} \sin(2 \pi V)$$
where $U$ and $V$ are independent uniform random variables over $[0,1]$.
I started solving it using the change of variables, so
$U = e^{\frac{-1}{2}(X^2 +Y^2)}$ and $ V = \frac{1}{2 \pi}\operatorname{atan2}(Y,\,X)$.
then using formula of the joint distribution $$f_{X,Y}(x,y)=|J| f(u(x,y),v(x,y))$$ I should obtain the joint as $\frac{1}{2 \pi} e^{\frac{-1}{2} (x^2 +y^2)}$
while what I found is totally different. Could anyone help me please in this last step.
What I found
$ f(u,v)= 1$ since both variables are uniform and independent.
then the the determinant of Jacobian is $$ - \frac{x^2}{2 \pi(x^2+y^2)} e^{\frac{-1}{2}(x^2+y^2)} + \frac{y^2}{2 \pi(x^2+y^2)} e^{\frac{-1}{2}(x^2+y^2)}$$ which can't be simplified to the normal.
| \begin{align}
x & = \sqrt{-2\log u}\, \cos(2\pi v) \\[8pt]
y & = \sqrt{-2\log u}\, \sin(2\pi v)
\end{align}
\begin{align}
& \frac{\partial x}{\partial u} = \frac{-\cos(2\pi v)}{u\sqrt{-2\log u}} & & \frac{\partial y}{\partial u} = \frac{-\sin(2\pi v)}{u\sqrt{-2\log u}} \\[12pt]
& \frac{\partial x}{\partial v} = \sqrt{-2\log u}\,\sin(2\pi v) & & \frac{\partial y}{\partial v} = \sqrt{-2\log u}\,\cos(2\pi v)
\end{align}
Therefore
$$
\left| \frac{\partial(x,y)}{\partial(u,v)} \right| = \frac 1 u.
$$
So $ dx\,dy = \dfrac{du\,dv} u.$
$$
x^2 + y^2 = -2\log u
$$
$$
- \frac{x^2+y^2} 2 = \log u
$$
$$
e^{-(x^2+y^2)/2} = u
$$
$$
e^{-(x^2+y^2)/2}\,dx\,dy = u\cdot\frac{du\,dv} u = du\,dv
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve the recurrence relation $a_n - 5a_{n-1} +8a_{n-2} -4a_{n-3} = 3^n$ Solve the recurrence
$a_n - 5a_{n-1} +8a_{n-2} -4a_{n-3} = 3^n$
with initial conditions $a_1 = 1, a_2 =5$ and $a_3 = 17$ How this was calculated? any hint or idea highly appreciated
| Let $a_n=b_n \, 3^n$ to make
$$b_n-\frac{5}{3} b_{n-1}+\frac{8}{9} b_{n-2}-\frac{4}{27} b_{n-3}=1$$ Now, make $b_n=c_n+k$ to make
$$c_n-\frac{5}{3} c_{n-1}+\frac{8}{9} c_{n-2}-\frac{4}{27} c_{n-3}=1-\frac{2 k}{27}$$ Chose $k=\frac{27}{2}$ to make the rhs equal to $0$.
Solve
$$c_n-\frac{5}{3} c_{n-1}+\frac{8}{9} c_{n-2}-\frac{4}{27} c_{n-3}=0$$ When done, go back to $b_n=c_n+\frac{27}{2}$ and to $a_n=b_n \, 3^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $n^4 \mod 8$ is identically equal to either 0 or 1, $\forall \ n\in \mathbb{N}$. I feel pretty confident about the first half of my solution for this, however I don't like how I used the induction hypothesis on the case for even integers, it feels like it isn't doing anything useful since it is really easy to show that $P(2x+2) \mod 8 \equiv 0$ without using the inductive hypothesis.
This is also the first semester I've been doing proofs, so am I going about this right?
Proof. Since $1^4 \mod 8 = 1$, $P(1)$ holds. Also, $2^4 \mod 8 = 16 \mod 8 = 0$ so, $P(2)$ holds.
We claim that $P(2x-1)$ holds, then $(2x-1)^4 \mod 8\equiv 1$ for any odd integer $n=2x-1$. So,
$$(2x-1)^4 = 16x^4-32x^3+24x^2-8x+1 \mod 8\equiv 1,$$
now consider
$$(2x+1)^4 = 16x^4+32x^3+24x^2+8x+1 = (16x^4-32x^3+24x^2-8x+1)+64x^3+16x,$$
then since
$$64x^3+16x \mod 8 = 8(8x^3+2x) \mod 8 \equiv 0,$$
we conclude that
$$(16x^4-32x^3+24x^2-8x+1)+64x^3+16x \mod 8 \equiv 1+0 = 1.$$
Thus, $P(2x+1)$ holds, and all odd integers $n$ hold.
Next, we claim that $P(2x)$ holds, then $(2x)^4 \mod 8\equiv 0$ for any even integer $n=2x$. So,
$$(2x)^4 \mod 8 = 16x^4 \mod 8 \equiv 0,$$
now consider
$$(2x+2)^4 = 16x^4+64x^3+96x^2+64x+16,$$
then
$$64x^3+96x^2+64x+16 \mod 8 = 8(8x^3+12x^2+8x+2) \mod 8 \equiv 0,$$
we conclude that
$$16x^4+64x^3+96x^2+64x+16 \mod 8 \equiv 0+0 = 0.$$
Thus, $P(2x)$ holds, and all even integers $n$ hold.
By mathematical induction, if $n=2x-1$ and $n=2x$ hold for the given statement, $n=2x+1$ and $n=2x+2$ also hold. Therefore, the statement holds for all $n \in \mathbb{N}$.
$\mathbb{QED}$
| Your proof is unnecessarily complicated.
If $n$ is even, then we can say $n = 2k$
$(2k)^4 = 16k^4 = 8(2k^4)$ which is divisible by $8.$
$(2k)^4 \equiv 0 \pmod 8$
If $n$ is odd, then $n = 2k+1$
$(2k+1)^4 = 16k^4 + 4(2k)^3 + 6(2k)^2 + 4(2k)+ 1$
Each of the first $4$ terms is divisible by $8$
$(2k+1)^4 \equiv 1 \pmod 8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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